While extracting character from a string in C++, why do I have to do this?
string s; string adder;
for (int i = s.size ()-1; i >= 0; i--) {
adder += s[i];
}
cout << adder << endl;
I am trying to reverse a string using c++ and am confused why do we have to do s.size()-1 and why does it print a space when we don't have -1?
c++
edited Nov 21 '18 at 14:20
PC Luddite
4,71051536
add a comment |
string s; string adder;
for (int i = s.size ()-1; i >= 0; i--) {
adder += s[i];
}
cout << adder << endl;
I am trying to reverse a string using c++ and am confused why do we have to do s.size()-1 and why does it print a space when we don't have -1?
c++
edited Nov 21 '18 at 14:20
PC Luddite
4,71051536
5
Array indexes reside in[0, s.size()).s[s.size()]is one past the end of the array so you need to start ats.size()-1.
– 0x499602D2
Nov 21 '18 at 14:13
@0x499602D2 - That's an answer. So, you know...
– StoryTeller
Nov 21 '18 at 14:16
Possible duplicate of Why does the indexing start with zero in 'C'?
– dandan78
Nov 23 '18 at 13:45
add a comment |
string s; string adder;
for (int i = s.size ()-1; i >= 0; i--) {
adder += s[i];
}
cout << adder << endl;
I am trying to reverse a string using c++ and am confused why do we have to do s.size()-1 and why does it print a space when we don't have -1?
c++
edited Nov 21 '18 at 14:20
PC Luddite
4,71051536
string s; string adder;
for (int i = s.size ()-1; i >= 0; i--) {
adder += s[i];
}
cout << adder << endl;
I am trying to reverse a string using c++ and am confused why do we have to do s.size()-1 and why does it print a space when we don't have -1?
c++
c++
edited Nov 21 '18 at 14:20
PC Luddite
4,71051536
edited Nov 21 '18 at 14:20
PC Luddite
4,71051536
edited Nov 21 '18 at 14:20
PC Luddite
4,71051536
edited Nov 21 '18 at 14:20
PC Luddite
4,71051536
edited Nov 21 '18 at 14:20
PC Luddite
4,71051536
4,71051536
asked Nov 21 '18 at 14:11
user10655470
5
Array indexes reside in[0, s.size()).s[s.size()]is one past the end of the array so you need to start ats.size()-1.
– 0x499602D2
Nov 21 '18 at 14:13
@0x499602D2 - That's an answer. So, you know...
– StoryTeller
Nov 21 '18 at 14:16
Possible duplicate of Why does the indexing start with zero in 'C'?
– dandan78
Nov 23 '18 at 13:45
add a comment |
5
Array indexes reside in[0, s.size()).s[s.size()]is one past the end of the array so you need to start ats.size()-1.
– 0x499602D2
Nov 21 '18 at 14:13
@0x499602D2 - That's an answer. So, you know...
– StoryTeller
Nov 21 '18 at 14:16
Possible duplicate of Why does the indexing start with zero in 'C'?
– dandan78
Nov 23 '18 at 13:45
5
5
Array indexes reside in
[0, s.size()). s[s.size()] is one past the end of the array so you need to start at s.size()-1.– 0x499602D2
Nov 21 '18 at 14:13
Array indexes reside in
[0, s.size()). s[s.size()] is one past the end of the array so you need to start at s.size()-1.– 0x499602D2
Nov 21 '18 at 14:13
@0x499602D2 - That's an answer. So, you know...
– StoryTeller
Nov 21 '18 at 14:16
@0x499602D2 - That's an answer. So, you know...
– StoryTeller
Nov 21 '18 at 14:16
Possible duplicate of Why does the indexing start with zero in 'C'?
– dandan78
Nov 23 '18 at 13:45
Possible duplicate of Why does the indexing start with zero in 'C'?
– dandan78
Nov 23 '18 at 13:45
add a comment |
3 Answers
3
active
oldest
votes
Array indexes reside in [0, s.size()). s[s.size()] is one past the end of the array so you need to start at s.size()-1.
You can see that this is needed if you use the at() member function which uses bounds checking:
adder += s.at(i); // throws exception if i is out of bounds
answered Nov 21 '18 at 14:16
0x499602D20x499602D2
68.1k26120207
add a comment |
In you code s is a string variable that holds some characters. And as you said you want to display its contents in reverse order.
In order to do that, you have to navigate through the contents of s from back to front. To do that, as shown in your code you used for-loop.
Let's say s contains 10 characters. Since you are accessing your string s back-to-front, the last character found at the 10 - 1 = 9th index, because it starts counting from 0, not 1.
EDIT with Example
string original = "Hello";
string reverse;
for (int i = original.size() - 1; i >= 0; i--) {
reverse += original[i];
}
cout << reverse << endl;
As you can see in the above example, original is a string variable which holds five characters, and reverse is also a string variable which is about to hold the contents of original in reverse order. In order to reverse the contents of original, we must navigate through it back-to-front.
original[4] = o
original[3] = l
original[2] = l
original[1] = e
original[0] = H
The above characters will be added one by one at each iteration of the for-loop as shown below:
reverse = ""; //it was empty first
reverse = reverse + original[4]; // now it holds the last character -> o
reverse = reverse + original[3]; // now it holds the characters -> ol
reverse = reverse + original[2]; // now it holds the characters -> oll
reverse = reverse + original[1]; // now it holds the characters -> olle
reverse = reverse + original[0]; // now it holds the characters -> olleh
edited Nov 23 '18 at 13:42
rioki
4,07942347
answered Nov 21 '18 at 14:18
Hello WorldHello World
390212
While allocating an array let's say int array[5]. So this array can store 6 elements then?
– user10655470
Nov 21 '18 at 19:04
@CyborgGamer No it stores 5 elements, and the indexes go from0to4.
– 0x499602D2
Nov 21 '18 at 22:33
@CyborgGamer No it is not. The confusion will banish once you are more familiar witharrays. Array of size 5 will contains only 5 elements (array[0], array[1], array[2], array[3] and array[4]).
– Hello World
Nov 22 '18 at 10:37
Hello guys, I am really rusty on arrays. Especially in C. How can I improve my knowledge on array?
– user10655470
Nov 23 '18 at 11:04
1
@CyborgGamer you should try to understand arrays from the scratch. Just ignore using your question as an example. If an array has size of 6, then it will always contain 6 items in it. But instead of accessing the last item usingarray[6], we will usearray[5], because the counting starts fromarray[0]instead ofarray[1].array[0]hold the first item,array[1]the second, ... andarray[5]the sixth. It is simple as that.
– Hello World
Dec 3 '18 at 5:58
|
show 1 more comment
It is because indexing is zero-based. So the first element is at position zero.
Since you use c++ you may (and should) use reverse iterators:
for (auto it = str.rbegin(); it != str.rend(); ++it) {
char& c = *it;
//do stuff
}
If you have boost you can just do:
for (auto c : boost::adaptors::reversed(str)) {
}
answered Nov 21 '18 at 14:21
darunedarune
1,615617
How does this answer the OP? And how does it help OP understand zero-based indexes?
– Jim Rhodes
Nov 21 '18 at 14:23
Yup, this should be a comment.
– HolyBlackCat
Nov 21 '18 at 14:24
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Array indexes reside in [0, s.size()). s[s.size()] is one past the end of the array so you need to start at s.size()-1.
You can see that this is needed if you use the at() member function which uses bounds checking:
adder += s.at(i); // throws exception if i is out of bounds
answered Nov 21 '18 at 14:16
0x499602D20x499602D2
68.1k26120207
add a comment |
Array indexes reside in [0, s.size()). s[s.size()] is one past the end of the array so you need to start at s.size()-1.
You can see that this is needed if you use the at() member function which uses bounds checking:
adder += s.at(i); // throws exception if i is out of bounds
answered Nov 21 '18 at 14:16
0x499602D20x499602D2
68.1k26120207
add a comment |
Array indexes reside in [0, s.size()). s[s.size()] is one past the end of the array so you need to start at s.size()-1.
You can see that this is needed if you use the at() member function which uses bounds checking:
adder += s.at(i); // throws exception if i is out of bounds
answered Nov 21 '18 at 14:16
0x499602D20x499602D2
68.1k26120207
Array indexes reside in [0, s.size()). s[s.size()] is one past the end of the array so you need to start at s.size()-1.
You can see that this is needed if you use the at() member function which uses bounds checking:
adder += s.at(i); // throws exception if i is out of bounds
answered Nov 21 '18 at 14:16
0x499602D20x499602D2
68.1k26120207
edited Nov 21 '18 at 14:23
answered Nov 21 '18 at 14:16
0x499602D20x499602D2
68.1k26120207
answered Nov 21 '18 at 14:16
0x499602D20x499602D2
68.1k26120207
answered Nov 21 '18 at 14:16
0x499602D20x499602D2
68.1k26120207
68.1k26120207
add a comment |
add a comment |
In you code s is a string variable that holds some characters. And as you said you want to display its contents in reverse order.
In order to do that, you have to navigate through the contents of s from back to front. To do that, as shown in your code you used for-loop.
Let's say s contains 10 characters. Since you are accessing your string s back-to-front, the last character found at the 10 - 1 = 9th index, because it starts counting from 0, not 1.
EDIT with Example
string original = "Hello";
string reverse;
for (int i = original.size() - 1; i >= 0; i--) {
reverse += original[i];
}
cout << reverse << endl;
As you can see in the above example, original is a string variable which holds five characters, and reverse is also a string variable which is about to hold the contents of original in reverse order. In order to reverse the contents of original, we must navigate through it back-to-front.
original[4] = o
original[3] = l
original[2] = l
original[1] = e
original[0] = H
The above characters will be added one by one at each iteration of the for-loop as shown below:
reverse = ""; //it was empty first
reverse = reverse + original[4]; // now it holds the last character -> o
reverse = reverse + original[3]; // now it holds the characters -> ol
reverse = reverse + original[2]; // now it holds the characters -> oll
reverse = reverse + original[1]; // now it holds the characters -> olle
reverse = reverse + original[0]; // now it holds the characters -> olleh
edited Nov 23 '18 at 13:42
rioki
4,07942347
answered Nov 21 '18 at 14:18
Hello WorldHello World
390212
While allocating an array let's say int array[5]. So this array can store 6 elements then?
– user10655470
Nov 21 '18 at 19:04
@CyborgGamer No it stores 5 elements, and the indexes go from0to4.
– 0x499602D2
Nov 21 '18 at 22:33
@CyborgGamer No it is not. The confusion will banish once you are more familiar witharrays. Array of size 5 will contains only 5 elements (array[0], array[1], array[2], array[3] and array[4]).
– Hello World
Nov 22 '18 at 10:37
Hello guys, I am really rusty on arrays. Especially in C. How can I improve my knowledge on array?
– user10655470
Nov 23 '18 at 11:04
1
@CyborgGamer you should try to understand arrays from the scratch. Just ignore using your question as an example. If an array has size of 6, then it will always contain 6 items in it. But instead of accessing the last item usingarray[6], we will usearray[5], because the counting starts fromarray[0]instead ofarray[1].array[0]hold the first item,array[1]the second, ... andarray[5]the sixth. It is simple as that.
– Hello World
Dec 3 '18 at 5:58
|
show 1 more comment
In you code s is a string variable that holds some characters. And as you said you want to display its contents in reverse order.
In order to do that, you have to navigate through the contents of s from back to front. To do that, as shown in your code you used for-loop.
Let's say s contains 10 characters. Since you are accessing your string s back-to-front, the last character found at the 10 - 1 = 9th index, because it starts counting from 0, not 1.
EDIT with Example
string original = "Hello";
string reverse;
for (int i = original.size() - 1; i >= 0; i--) {
reverse += original[i];
}
cout << reverse << endl;
As you can see in the above example, original is a string variable which holds five characters, and reverse is also a string variable which is about to hold the contents of original in reverse order. In order to reverse the contents of original, we must navigate through it back-to-front.
original[4] = o
original[3] = l
original[2] = l
original[1] = e
original[0] = H
The above characters will be added one by one at each iteration of the for-loop as shown below:
reverse = ""; //it was empty first
reverse = reverse + original[4]; // now it holds the last character -> o
reverse = reverse + original[3]; // now it holds the characters -> ol
reverse = reverse + original[2]; // now it holds the characters -> oll
reverse = reverse + original[1]; // now it holds the characters -> olle
reverse = reverse + original[0]; // now it holds the characters -> olleh
edited Nov 23 '18 at 13:42
rioki
4,07942347
answered Nov 21 '18 at 14:18
Hello WorldHello World
390212
While allocating an array let's say int array[5]. So this array can store 6 elements then?
– user10655470
Nov 21 '18 at 19:04
@CyborgGamer No it stores 5 elements, and the indexes go from0to4.
– 0x499602D2
Nov 21 '18 at 22:33
@CyborgGamer No it is not. The confusion will banish once you are more familiar witharrays. Array of size 5 will contains only 5 elements (array[0], array[1], array[2], array[3] and array[4]).
– Hello World
Nov 22 '18 at 10:37
Hello guys, I am really rusty on arrays. Especially in C. How can I improve my knowledge on array?
– user10655470
Nov 23 '18 at 11:04
1
@CyborgGamer you should try to understand arrays from the scratch. Just ignore using your question as an example. If an array has size of 6, then it will always contain 6 items in it. But instead of accessing the last item usingarray[6], we will usearray[5], because the counting starts fromarray[0]instead ofarray[1].array[0]hold the first item,array[1]the second, ... andarray[5]the sixth. It is simple as that.
– Hello World
Dec 3 '18 at 5:58
|
show 1 more comment
In you code s is a string variable that holds some characters. And as you said you want to display its contents in reverse order.
In order to do that, you have to navigate through the contents of s from back to front. To do that, as shown in your code you used for-loop.
Let's say s contains 10 characters. Since you are accessing your string s back-to-front, the last character found at the 10 - 1 = 9th index, because it starts counting from 0, not 1.
EDIT with Example
string original = "Hello";
string reverse;
for (int i = original.size() - 1; i >= 0; i--) {
reverse += original[i];
}
cout << reverse << endl;
As you can see in the above example, original is a string variable which holds five characters, and reverse is also a string variable which is about to hold the contents of original in reverse order. In order to reverse the contents of original, we must navigate through it back-to-front.
original[4] = o
original[3] = l
original[2] = l
original[1] = e
original[0] = H
The above characters will be added one by one at each iteration of the for-loop as shown below:
reverse = ""; //it was empty first
reverse = reverse + original[4]; // now it holds the last character -> o
reverse = reverse + original[3]; // now it holds the characters -> ol
reverse = reverse + original[2]; // now it holds the characters -> oll
reverse = reverse + original[1]; // now it holds the characters -> olle
reverse = reverse + original[0]; // now it holds the characters -> olleh
edited Nov 23 '18 at 13:42
rioki
4,07942347
answered Nov 21 '18 at 14:18
Hello WorldHello World
390212
In you code s is a string variable that holds some characters. And as you said you want to display its contents in reverse order.
In order to do that, you have to navigate through the contents of s from back to front. To do that, as shown in your code you used for-loop.
Let's say s contains 10 characters. Since you are accessing your string s back-to-front, the last character found at the 10 - 1 = 9th index, because it starts counting from 0, not 1.
EDIT with Example
string original = "Hello";
string reverse;
for (int i = original.size() - 1; i >= 0; i--) {
reverse += original[i];
}
cout << reverse << endl;
As you can see in the above example, original is a string variable which holds five characters, and reverse is also a string variable which is about to hold the contents of original in reverse order. In order to reverse the contents of original, we must navigate through it back-to-front.
original[4] = o
original[3] = l
original[2] = l
original[1] = e
original[0] = H
The above characters will be added one by one at each iteration of the for-loop as shown below:
reverse = ""; //it was empty first
reverse = reverse + original[4]; // now it holds the last character -> o
reverse = reverse + original[3]; // now it holds the characters -> ol
reverse = reverse + original[2]; // now it holds the characters -> oll
reverse = reverse + original[1]; // now it holds the characters -> olle
reverse = reverse + original[0]; // now it holds the characters -> olleh
edited Nov 23 '18 at 13:42
rioki
4,07942347
answered Nov 21 '18 at 14:18
Hello WorldHello World
390212
edited Nov 23 '18 at 13:42
rioki
4,07942347
edited Nov 23 '18 at 13:42
rioki
4,07942347
edited Nov 23 '18 at 13:42
rioki
4,07942347
4,07942347
answered Nov 21 '18 at 14:18
Hello WorldHello World
390212
answered Nov 21 '18 at 14:18
Hello WorldHello World
390212
answered Nov 21 '18 at 14:18
Hello WorldHello World
390212
390212
While allocating an array let's say int array[5]. So this array can store 6 elements then?
– user10655470
Nov 21 '18 at 19:04
@CyborgGamer No it stores 5 elements, and the indexes go from0to4.
– 0x499602D2
Nov 21 '18 at 22:33
@CyborgGamer No it is not. The confusion will banish once you are more familiar witharrays. Array of size 5 will contains only 5 elements (array[0], array[1], array[2], array[3] and array[4]).
– Hello World
Nov 22 '18 at 10:37
Hello guys, I am really rusty on arrays. Especially in C. How can I improve my knowledge on array?
– user10655470
Nov 23 '18 at 11:04
1
@CyborgGamer you should try to understand arrays from the scratch. Just ignore using your question as an example. If an array has size of 6, then it will always contain 6 items in it. But instead of accessing the last item usingarray[6], we will usearray[5], because the counting starts fromarray[0]instead ofarray[1].array[0]hold the first item,array[1]the second, ... andarray[5]the sixth. It is simple as that.
– Hello World
Dec 3 '18 at 5:58
|
show 1 more comment
While allocating an array let's say int array[5]. So this array can store 6 elements then?
– user10655470
Nov 21 '18 at 19:04
@CyborgGamer No it stores 5 elements, and the indexes go from0to4.
– 0x499602D2
Nov 21 '18 at 22:33
@CyborgGamer No it is not. The confusion will banish once you are more familiar witharrays. Array of size 5 will contains only 5 elements (array[0], array[1], array[2], array[3] and array[4]).
– Hello World
Nov 22 '18 at 10:37
Hello guys, I am really rusty on arrays. Especially in C. How can I improve my knowledge on array?
– user10655470
Nov 23 '18 at 11:04
1
@CyborgGamer you should try to understand arrays from the scratch. Just ignore using your question as an example. If an array has size of 6, then it will always contain 6 items in it. But instead of accessing the last item usingarray[6], we will usearray[5], because the counting starts fromarray[0]instead ofarray[1].array[0]hold the first item,array[1]the second, ... andarray[5]the sixth. It is simple as that.
– Hello World
Dec 3 '18 at 5:58
While allocating an array let's say int array[5]. So this array can store 6 elements then?
– user10655470
Nov 21 '18 at 19:04
While allocating an array let's say int array[5]. So this array can store 6 elements then?
– user10655470
Nov 21 '18 at 19:04
@CyborgGamer No it stores 5 elements, and the indexes go from
0 to 4.– 0x499602D2
Nov 21 '18 at 22:33
@CyborgGamer No it stores 5 elements, and the indexes go from
0 to 4.– 0x499602D2
Nov 21 '18 at 22:33
@CyborgGamer No it is not. The confusion will banish once you are more familiar with
arrays. Array of size 5 will contains only 5 elements (array[0], array[1], array[2], array[3] and array[4]).– Hello World
Nov 22 '18 at 10:37
@CyborgGamer No it is not. The confusion will banish once you are more familiar with
arrays. Array of size 5 will contains only 5 elements (array[0], array[1], array[2], array[3] and array[4]).– Hello World
Nov 22 '18 at 10:37
Hello guys, I am really rusty on arrays. Especially in C. How can I improve my knowledge on array?
– user10655470
Nov 23 '18 at 11:04
Hello guys, I am really rusty on arrays. Especially in C. How can I improve my knowledge on array?
– user10655470
Nov 23 '18 at 11:04
1
1
@CyborgGamer you should try to understand arrays from the scratch. Just ignore using your question as an example. If an array has size of 6, then it will always contain 6 items in it. But instead of accessing the last item using
array[6], we will use array[5], because the counting starts from array[0] instead of array[1]. array[0] hold the first item, array[1] the second, ... and array[5] the sixth. It is simple as that.– Hello World
Dec 3 '18 at 5:58
@CyborgGamer you should try to understand arrays from the scratch. Just ignore using your question as an example. If an array has size of 6, then it will always contain 6 items in it. But instead of accessing the last item using
array[6], we will use array[5], because the counting starts from array[0] instead of array[1]. array[0] hold the first item, array[1] the second, ... and array[5] the sixth. It is simple as that.– Hello World
Dec 3 '18 at 5:58
|
show 1 more comment
It is because indexing is zero-based. So the first element is at position zero.
Since you use c++ you may (and should) use reverse iterators:
for (auto it = str.rbegin(); it != str.rend(); ++it) {
char& c = *it;
//do stuff
}
If you have boost you can just do:
for (auto c : boost::adaptors::reversed(str)) {
}
answered Nov 21 '18 at 14:21
darunedarune
1,615617
How does this answer the OP? And how does it help OP understand zero-based indexes?
– Jim Rhodes
Nov 21 '18 at 14:23
Yup, this should be a comment.
– HolyBlackCat
Nov 21 '18 at 14:24
add a comment |
It is because indexing is zero-based. So the first element is at position zero.
Since you use c++ you may (and should) use reverse iterators:
for (auto it = str.rbegin(); it != str.rend(); ++it) {
char& c = *it;
//do stuff
}
If you have boost you can just do:
for (auto c : boost::adaptors::reversed(str)) {
}
answered Nov 21 '18 at 14:21
darunedarune
1,615617
How does this answer the OP? And how does it help OP understand zero-based indexes?
– Jim Rhodes
Nov 21 '18 at 14:23
Yup, this should be a comment.
– HolyBlackCat
Nov 21 '18 at 14:24
add a comment |
It is because indexing is zero-based. So the first element is at position zero.
Since you use c++ you may (and should) use reverse iterators:
for (auto it = str.rbegin(); it != str.rend(); ++it) {
char& c = *it;
//do stuff
}
If you have boost you can just do:
for (auto c : boost::adaptors::reversed(str)) {
}
answered Nov 21 '18 at 14:21
darunedarune
1,615617
It is because indexing is zero-based. So the first element is at position zero.
Since you use c++ you may (and should) use reverse iterators:
for (auto it = str.rbegin(); it != str.rend(); ++it) {
char& c = *it;
//do stuff
}
If you have boost you can just do:
for (auto c : boost::adaptors::reversed(str)) {
}
answered Nov 21 '18 at 14:21
darunedarune
1,615617
edited Nov 21 '18 at 14:30
answered Nov 21 '18 at 14:21
darunedarune
1,615617
answered Nov 21 '18 at 14:21
darunedarune
1,615617
answered Nov 21 '18 at 14:21
darunedarune
1,615617
1,615617
How does this answer the OP? And how does it help OP understand zero-based indexes?
– Jim Rhodes
Nov 21 '18 at 14:23
Yup, this should be a comment.
– HolyBlackCat
Nov 21 '18 at 14:24
add a comment |
How does this answer the OP? And how does it help OP understand zero-based indexes?
– Jim Rhodes
Nov 21 '18 at 14:23
Yup, this should be a comment.
– HolyBlackCat
Nov 21 '18 at 14:24
How does this answer the OP? And how does it help OP understand zero-based indexes?
– Jim Rhodes
Nov 21 '18 at 14:23
How does this answer the OP? And how does it help OP understand zero-based indexes?
– Jim Rhodes
Nov 21 '18 at 14:23
Yup, this should be a comment.
– HolyBlackCat
Nov 21 '18 at 14:24
Yup, this should be a comment.
– HolyBlackCat
Nov 21 '18 at 14:24
add a comment |
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5
Array indexes reside in
[0, s.size()).s[s.size()]is one past the end of the array so you need to start ats.size()-1.– 0x499602D2
Nov 21 '18 at 14:13
@0x499602D2 - That's an answer. So, you know...
– StoryTeller
Nov 21 '18 at 14:16
Possible duplicate of Why does the indexing start with zero in 'C'?
– dandan78
Nov 23 '18 at 13:45