scanf only reads first field

I want to read in two strings with scanf and store them in two variables. How ever only the first one seems to be read properly, the second one just returns (null), but I'm not sure why though.

int main(int argc, char **argv) {

  char *c, *p;



  printf("Enter a command with a parameter: ");

  scanf("%s%s", c, p);

  ...

}

Somebody got an idea, what's going wrong? I don't find any errors and for my understanding it should work.

asked Nov 19 '18 at 15:34

anonymoose

335

When I enter ls -l it only reads the ls properly, that's what i don't understand. scanf should read the white space between them and store the -l in p, but i doesn't.

– anonymoose
Nov 19 '18 at 15:39

3

You've not initialized either c or p to point to anywhere, so you have undefined behaviour.

– Jonathan Leffler
Nov 19 '18 at 15:42

It’s undefined behavior you didn’t initialize the char * variables.

– danglingpointer
Nov 19 '18 at 15:42

You don't look at the return value of scanf. How do you know how many fields were parsed?

– Gerhardh
Nov 19 '18 at 15:50

FYI: There's a Q&A How to prevent scanf() causing buffer overflow in C?

– Jonathan Leffler
Nov 19 '18 at 15:54

add a comment |

I want to read in two strings with scanf and store them in two variables. How ever only the first one seems to be read properly, the second one just returns (null), but I'm not sure why though.

int main(int argc, char **argv) {

  char *c, *p;



  printf("Enter a command with a parameter: ");

  scanf("%s%s", c, p);

  ...

}

Somebody got an idea, what's going wrong? I don't find any errors and for my understanding it should work.

asked Nov 19 '18 at 15:34

anonymoose

335

When I enter ls -l it only reads the ls properly, that's what i don't understand. scanf should read the white space between them and store the -l in p, but i doesn't.

– anonymoose
Nov 19 '18 at 15:39

3

You've not initialized either c or p to point to anywhere, so you have undefined behaviour.

– Jonathan Leffler
Nov 19 '18 at 15:42

It’s undefined behavior you didn’t initialize the char * variables.

– danglingpointer
Nov 19 '18 at 15:42

You don't look at the return value of scanf. How do you know how many fields were parsed?

– Gerhardh
Nov 19 '18 at 15:50

FYI: There's a Q&A How to prevent scanf() causing buffer overflow in C?

– Jonathan Leffler
Nov 19 '18 at 15:54

add a comment |

I want to read in two strings with scanf and store them in two variables. How ever only the first one seems to be read properly, the second one just returns (null), but I'm not sure why though.

int main(int argc, char **argv) {

  char *c, *p;



  printf("Enter a command with a parameter: ");

  scanf("%s%s", c, p);

  ...

}

Somebody got an idea, what's going wrong? I don't find any errors and for my understanding it should work.

asked Nov 19 '18 at 15:34

anonymoose

335

I want to read in two strings with scanf and store them in two variables. How ever only the first one seems to be read properly, the second one just returns (null), but I'm not sure why though.

int main(int argc, char **argv) {

  char *c, *p;



  printf("Enter a command with a parameter: ");

  scanf("%s%s", c, p);

  ...

}

Somebody got an idea, what's going wrong? I don't find any errors and for my understanding it should work.

c scanf stdio

asked Nov 19 '18 at 15:34

anonymoose

335

asked Nov 19 '18 at 15:34

anonymoose

335

asked Nov 19 '18 at 15:34

anonymoose

335

asked Nov 19 '18 at 15:34

anonymoose

335

asked Nov 19 '18 at 15:34

anonymoose

335

When I enter ls -l it only reads the ls properly, that's what i don't understand. scanf should read the white space between them and store the -l in p, but i doesn't.

– anonymoose
Nov 19 '18 at 15:39

3

You've not initialized either c or p to point to anywhere, so you have undefined behaviour.

– Jonathan Leffler
Nov 19 '18 at 15:42

It’s undefined behavior you didn’t initialize the char * variables.

– danglingpointer
Nov 19 '18 at 15:42

You don't look at the return value of scanf. How do you know how many fields were parsed?

– Gerhardh
Nov 19 '18 at 15:50

FYI: There's a Q&A How to prevent scanf() causing buffer overflow in C?

– Jonathan Leffler
Nov 19 '18 at 15:54

add a comment |

When I enter ls -l it only reads the ls properly, that's what i don't understand. scanf should read the white space between them and store the -l in p, but i doesn't.

– anonymoose
Nov 19 '18 at 15:39

3

You've not initialized either c or p to point to anywhere, so you have undefined behaviour.

– Jonathan Leffler
Nov 19 '18 at 15:42

It’s undefined behavior you didn’t initialize the char * variables.

– danglingpointer
Nov 19 '18 at 15:42

You don't look at the return value of scanf. How do you know how many fields were parsed?

– Gerhardh
Nov 19 '18 at 15:50

FYI: There's a Q&A How to prevent scanf() causing buffer overflow in C?

– Jonathan Leffler
Nov 19 '18 at 15:54

When I enter ls -l it only reads the ls properly, that's what i don't understand. scanf should read the white space between them and store the -l in p, but i doesn't.

– anonymoose
Nov 19 '18 at 15:39

You've not initialized either c or p to point to anywhere, so you have undefined behaviour.

– Jonathan Leffler
Nov 19 '18 at 15:42

It’s undefined behavior you didn’t initialize the char * variables.

– danglingpointer
Nov 19 '18 at 15:42

You don't look at the return value of scanf. How do you know how many fields were parsed?

– Gerhardh
Nov 19 '18 at 15:50

FYI: There's a Q&A How to prevent scanf() causing buffer overflow in C?

– Jonathan Leffler
Nov 19 '18 at 15:54

add a comment |

3 Answers
3

active

oldest

votes

char *c, *p;

These are just pointers without any memory behind them. Moreover, they are uninitialized, so reading or assigning they value or reading from or assigning to the memory they point to would be undefined behavior and will spawn nasal demons.

In order to read input from user, you need a place to save the memory to.

char c[20], p[20];

Will allocate 20 characters for c and 20 character for p. c and p are not pointers, but arrays char[20]. However arrays handlers "decay" (read: magically change) into pointers in most contexts.

scanf for "%s" scanf modifier expects a pointer to char char* with enough memory to hold the inputted string.

I would do:

int main(int argc, char **argv) {

  char c[20], p[20];



  printf("Enter a command with a parameter: ");

  scanf("%19s%19s", c, p);

}

The 19 in the %19s limits scanf so that it does not reads input into memory region that is out of bounds in c and/or p pointers, which will lead to undefined behaviour too.

edited Nov 19 '18 at 15:49

answered Nov 19 '18 at 15:43

Kamil Cuk

9,9611527

1

Thanks a lot, i didn't even knew about the limitation stuff.

– anonymoose
Nov 19 '18 at 15:49

@JonathanLeffler Or just remove the comment, since it's no longer needed. :)

– Broman
Nov 19 '18 at 15:51

add a comment |

This is undefined behavior. You have not initialized your pointers. Your compiler should tell you about this.

k.c:8:17: warning: variable 'c' is uninitialized when used here [-Wuninitialized]

  scanf("%s%s", c, p);

What you need to do is to initialize your pointers. Using malloc is one way of doing it.

char *c=malloc(SIZE);

char *p=malloc(SIZE);

You want to avoid undefined behavior at ALL COST because it means that the behavior can be ANYTHING, including working the way you want. It can make debugging a nightmare.

But as Swordfish mentioned in the comments, it is suicidal to do this without specifying a max length for your string, because you will have no control over if you write past your buffer, which also is undefined behavior.

edited Nov 19 '18 at 15:46

answered Nov 19 '18 at 15:41

Broman

6,256112241

add a comment |

you can also use %ms instead of %s because it allocate memory itself and there is no need of allocation memory manually by programmer.

int main(int argc, char **argv) {

  char *c, *p;



  printf("Enter a command with a parameter: ");

  scanf("%ms%ms", &c, &p); //no need of allocating memory manually

  ... 

}

answered Nov 19 '18 at 16:51

Susheel Dwivedi

554

1

"%ms" is a useful idea with libraries that employ this non-standard enhancement. difference between %ms and %s scanf. Note that code still should later free(c), free(p)

– chux
Nov 19 '18 at 18:47

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

char *c, *p;

In order to read input from user, you need a place to save the memory to.

char c[20], p[20];

scanf for "%s" scanf modifier expects a pointer to char char* with enough memory to hold the inputted string.

I would do:

int main(int argc, char **argv) {

  char c[20], p[20];



  printf("Enter a command with a parameter: ");

  scanf("%19s%19s", c, p);

}

The 19 in the %19s limits scanf so that it does not reads input into memory region that is out of bounds in c and/or p pointers, which will lead to undefined behaviour too.

edited Nov 19 '18 at 15:49

answered Nov 19 '18 at 15:43

Kamil Cuk

9,9611527

1

Thanks a lot, i didn't even knew about the limitation stuff.

– anonymoose
Nov 19 '18 at 15:49

@JonathanLeffler Or just remove the comment, since it's no longer needed. :)

– Broman
Nov 19 '18 at 15:51

add a comment |

char *c, *p;

In order to read input from user, you need a place to save the memory to.

char c[20], p[20];

scanf for "%s" scanf modifier expects a pointer to char char* with enough memory to hold the inputted string.

I would do:

int main(int argc, char **argv) {

  char c[20], p[20];



  printf("Enter a command with a parameter: ");

  scanf("%19s%19s", c, p);

}

The 19 in the %19s limits scanf so that it does not reads input into memory region that is out of bounds in c and/or p pointers, which will lead to undefined behaviour too.

edited Nov 19 '18 at 15:49

answered Nov 19 '18 at 15:43

Kamil Cuk

9,9611527

1

Thanks a lot, i didn't even knew about the limitation stuff.

– anonymoose
Nov 19 '18 at 15:49

@JonathanLeffler Or just remove the comment, since it's no longer needed. :)

– Broman
Nov 19 '18 at 15:51

add a comment |

char *c, *p;

In order to read input from user, you need a place to save the memory to.

char c[20], p[20];

scanf for "%s" scanf modifier expects a pointer to char char* with enough memory to hold the inputted string.

I would do:

int main(int argc, char **argv) {

  char c[20], p[20];



  printf("Enter a command with a parameter: ");

  scanf("%19s%19s", c, p);

}

The 19 in the %19s limits scanf so that it does not reads input into memory region that is out of bounds in c and/or p pointers, which will lead to undefined behaviour too.

edited Nov 19 '18 at 15:49

answered Nov 19 '18 at 15:43

Kamil Cuk

9,9611527

char *c, *p;

In order to read input from user, you need a place to save the memory to.

char c[20], p[20];

scanf for "%s" scanf modifier expects a pointer to char char* with enough memory to hold the inputted string.

I would do:

int main(int argc, char **argv) {

  char c[20], p[20];



  printf("Enter a command with a parameter: ");

  scanf("%19s%19s", c, p);

}

The 19 in the %19s limits scanf so that it does not reads input into memory region that is out of bounds in c and/or p pointers, which will lead to undefined behaviour too.

edited Nov 19 '18 at 15:49

answered Nov 19 '18 at 15:43

Kamil Cuk

9,9611527

edited Nov 19 '18 at 15:49

answered Nov 19 '18 at 15:43

Kamil Cuk

9,9611527

answered Nov 19 '18 at 15:43

Kamil Cuk

9,9611527

answered Nov 19 '18 at 15:43

Kamil Cuk

9,9611527

1

Thanks a lot, i didn't even knew about the limitation stuff.

– anonymoose
Nov 19 '18 at 15:49

@JonathanLeffler Or just remove the comment, since it's no longer needed. :)

– Broman
Nov 19 '18 at 15:51

add a comment |

1

Thanks a lot, i didn't even knew about the limitation stuff.

– anonymoose
Nov 19 '18 at 15:49

@JonathanLeffler Or just remove the comment, since it's no longer needed. :)

– Broman
Nov 19 '18 at 15:51

Thanks a lot, i didn't even knew about the limitation stuff.

– anonymoose
Nov 19 '18 at 15:49

@JonathanLeffler Or just remove the comment, since it's no longer needed. :)

– Broman
Nov 19 '18 at 15:51

add a comment |

This is undefined behavior. You have not initialized your pointers. Your compiler should tell you about this.

k.c:8:17: warning: variable 'c' is uninitialized when used here [-Wuninitialized]

  scanf("%s%s", c, p);

What you need to do is to initialize your pointers. Using malloc is one way of doing it.

char *c=malloc(SIZE);

char *p=malloc(SIZE);

You want to avoid undefined behavior at ALL COST because it means that the behavior can be ANYTHING, including working the way you want. It can make debugging a nightmare.

edited Nov 19 '18 at 15:46

answered Nov 19 '18 at 15:41

Broman

6,256112241

add a comment |

This is undefined behavior. You have not initialized your pointers. Your compiler should tell you about this.

k.c:8:17: warning: variable 'c' is uninitialized when used here [-Wuninitialized]

  scanf("%s%s", c, p);

What you need to do is to initialize your pointers. Using malloc is one way of doing it.

char *c=malloc(SIZE);

char *p=malloc(SIZE);

You want to avoid undefined behavior at ALL COST because it means that the behavior can be ANYTHING, including working the way you want. It can make debugging a nightmare.

edited Nov 19 '18 at 15:46

answered Nov 19 '18 at 15:41

Broman

6,256112241

add a comment |

This is undefined behavior. You have not initialized your pointers. Your compiler should tell you about this.

k.c:8:17: warning: variable 'c' is uninitialized when used here [-Wuninitialized]

  scanf("%s%s", c, p);

What you need to do is to initialize your pointers. Using malloc is one way of doing it.

char *c=malloc(SIZE);

char *p=malloc(SIZE);

You want to avoid undefined behavior at ALL COST because it means that the behavior can be ANYTHING, including working the way you want. It can make debugging a nightmare.

edited Nov 19 '18 at 15:46

answered Nov 19 '18 at 15:41

Broman

6,256112241

This is undefined behavior. You have not initialized your pointers. Your compiler should tell you about this.

k.c:8:17: warning: variable 'c' is uninitialized when used here [-Wuninitialized]

  scanf("%s%s", c, p);

What you need to do is to initialize your pointers. Using malloc is one way of doing it.

char *c=malloc(SIZE);

char *p=malloc(SIZE);

You want to avoid undefined behavior at ALL COST because it means that the behavior can be ANYTHING, including working the way you want. It can make debugging a nightmare.

edited Nov 19 '18 at 15:46

answered Nov 19 '18 at 15:41

Broman

6,256112241

edited Nov 19 '18 at 15:46

answered Nov 19 '18 at 15:41

Broman

6,256112241

answered Nov 19 '18 at 15:41

Broman

6,256112241

answered Nov 19 '18 at 15:41

Broman

6,256112241

add a comment |

you can also use %ms instead of %s because it allocate memory itself and there is no need of allocation memory manually by programmer.

int main(int argc, char **argv) {

  char *c, *p;



  printf("Enter a command with a parameter: ");

  scanf("%ms%ms", &c, &p); //no need of allocating memory manually

  ... 

}

answered Nov 19 '18 at 16:51

Susheel Dwivedi

554

1

"%ms" is a useful idea with libraries that employ this non-standard enhancement. difference between %ms and %s scanf. Note that code still should later free(c), free(p)

– chux
Nov 19 '18 at 18:47

add a comment |

you can also use %ms instead of %s because it allocate memory itself and there is no need of allocation memory manually by programmer.

int main(int argc, char **argv) {

  char *c, *p;



  printf("Enter a command with a parameter: ");

  scanf("%ms%ms", &c, &p); //no need of allocating memory manually

  ... 

}

answered Nov 19 '18 at 16:51

Susheel Dwivedi

554

1

"%ms" is a useful idea with libraries that employ this non-standard enhancement. difference between %ms and %s scanf. Note that code still should later free(c), free(p)

– chux
Nov 19 '18 at 18:47

add a comment |

you can also use %ms instead of %s because it allocate memory itself and there is no need of allocation memory manually by programmer.

int main(int argc, char **argv) {

  char *c, *p;



  printf("Enter a command with a parameter: ");

  scanf("%ms%ms", &c, &p); //no need of allocating memory manually

  ... 

}

answered Nov 19 '18 at 16:51

Susheel Dwivedi

554

you can also use %ms instead of %s because it allocate memory itself and there is no need of allocation memory manually by programmer.

int main(int argc, char **argv) {

  char *c, *p;



  printf("Enter a command with a parameter: ");

  scanf("%ms%ms", &c, &p); //no need of allocating memory manually

  ... 

}

answered Nov 19 '18 at 16:51

Susheel Dwivedi

554

answered Nov 19 '18 at 16:51

Susheel Dwivedi

554

answered Nov 19 '18 at 16:51

Susheel Dwivedi

554

answered Nov 19 '18 at 16:51

Susheel Dwivedi

554

1

"%ms" is a useful idea with libraries that employ this non-standard enhancement. difference between %ms and %s scanf. Note that code still should later free(c), free(p)

– chux
Nov 19 '18 at 18:47

add a comment |

1

"%ms" is a useful idea with libraries that employ this non-standard enhancement. difference between %ms and %s scanf. Note that code still should later free(c), free(p)

– chux
Nov 19 '18 at 18:47

"%ms" is a useful idea with libraries that employ this non-standard enhancement. difference between %ms and %s scanf. Note that code still should later free(c), free(p)

– chux
Nov 19 '18 at 18:47

add a comment |

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