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I am new to C and learning structs. I am trying to malloc a char pointer with size 30 but it is giving a segmentation fault(core dump). I searched it on the internet & SO but am not able to resolve this. Any help will be much appreciated.
Probably I am accessing the char* member of the struct incorrectly ?

typedef struct{ int x; int y; char *f; char *l; }str; void create_mall(); void create_mall() //Malloc the struct { str *p; p->f = (char*)malloc(sizeof(char)*30); // segmentation fault here p->l = (char*)malloc(sizeof(char)*30); printf("Enter the user ID:"); scanf("%d",&p->x); printf("\nEnter the phone number:"); scanf("%d",&p->y); printf("\nEnter the First name:"); scanf("%29s",p->f); printf("\nEnter the Last name:"); scanf("%29s",p->l); printf("\nEntered values are: %d %d %s %s\n",p->x,p->y,p->f,p->l); } int main(void) { create_mall(); return 0; }
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  • Note that the declaration void create_mall(); simply announces the presence of a function called create_mall() that returns no value but which takes any (fixed) number of arguments of indeterminate type. This is quite different from void create_mall(void); which says there's a function called create_mall() that returns no value and takes no arguments. In other words, what you've provided is not a prototype for the function in C. (In C++, it would be a prototype for a function taking no arguments and returning no value, but the language tag is C, not C++.) Commented Oct 18, 2012 at 5:10
  • Congratulations. Your question / problem did appear in my programming exam (including your text). Commented Feb 2, 2016 at 23:31

5 Answers 5

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Here's your problem:

str *p;

You've declared a pointer to an instance of str, but you haven't initialized it with a value. You either need to move this variable to the stack:

str p;

...or malloc some memory for it first:

str *p = (str*)malloc(sizeof(str));
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6 Comments

Pretty please, don't advice beginners to cast the return value of malloc()!
Why not, exactly? It promotes explicit programming and keeping tabs on your variable types.
@AdamMaras Here's why. Basically it's unnecessary and it clutters code and decreases readability. This is C, not C++.
@H2CO3 I disagree entirely with your reasoning. This makes the code incompatible with C++ compilers and less explicit. And the argument about forgetting to #include <stdlib.h> is crap... if you can't have the foresight to include the right headers for your functions, your application has no business executing without errors.
@AdamMaras "And the argument about forgetting to #include <stdlib.h> is crap" <- no, it isn't. I sometimes forget to include the right headers - God bless the compiler that it warns me then. "This makes the code incompatible with C++ compilers " <- Any decent C++ compiler should have a C mode. For example, GCC deduces the language from the file extension, regardless of whether one uses gcc or g++. "Less explicit" <- What kind of explicitness are you expecting here? void * is implicitly compatible with any data pointer type.
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5

You never allocated space for the struct itself, only a pointer to it.

Try something like:

str *p = malloc(sizeof(str));
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8 Comments

+1 for being the only answer that didn't cast the malloc() result.
@WhozCraig why shouldn't we cast the result of malloc() ?
@mux That's why.
@H2CO3 thanks for posting the link. I was seconds from linking the same exact post =P
@PrototypeStark C is not stupid as opposed to C++.
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As many people have pointed out, you need to allocate memory for that str struct, before writing the fields of it.

The best way to do so in C is:

p = malloc(sizeof *p);

This has the following advantages:

  1. No cast, since no cast is needed in C and having a cast can hide actual errors.
  2. No duplication of type information, by using the sizeof operator to compute how much storage is needed for the value p points at.

When you then allocate the string space, you can simplify it to:

p->f = malloc(30);

Because:

  1. No cast, for the very same reason.
  2. C guarantees that sizeof (char) is always 1, so using it like you did adds nothing, 1 * 30 is always just 30.

Last, you should always check the return value of malloc() before using it, since it can fail and return NULL.

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Check for NULL values in return of malloc() function.

Also str *p; < is not initialised.

initialize p as str *p = malloc(sizeof(str));

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The problem lies here.

str *p; ---> Problem Line 1<br> p->f = (char*)malloc(sizeof(char)*30); ----> Problem Line2 p->l = (char*)malloc(sizeof(char)*30);

You have declared a pointer p of type str.
Problem 1:
You have not initialized this pointer to NULL. Thus, p can point to anything.
Problem 2:
Since p is an uninitialized pointer, p->f can point anywhere which is causing the segfault. Below is the correct way

str *p = NULL; p = malloc(sizeof(str)); // Check p for NULL memset(p, 0, sizeof(str));

Now you have an initialized memory pointed by p. You are now free to use it as you want.

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2 Comments

if you assign p to the return value of malloc, it's not necessary to initialize it to NULL beforehand.
Yes, but its always better to initialize pointers to NULL at the time of declaration. If a user misses the malloc, he'll be able to identify it quickly.

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