malloc and pointers

hi all

Code:

---

#include <stdio.h>
#include <stdlib.h>

main()
{
      char *aa;
   
      if((aa=malloc(100 *sizeof(char)))==NULL)
          {
                        printf("Alloctaing Error\n");
                        exit(-1);
          }
     
      printf("%d",sizeof(aa));
}

---

it give me 4 as output !!!!!!!!!!!!!
but

Code:

---

#include <stdio.h>
#include <stdlib.h>

main()
{
      char  aa[100];
         
      printf("%d",sizeof(aa));
}

---

gives me 100 !!!
so where is the problem ?
why in the first code when i used malloc it gives me 4 but in the second code give me the object aa size ??
thanks in advance

In the first case aa is a char pointer. On 32bit machines pointers are 32bit = 4bytes, so that's why sizeof(aa) is returning 4.

In the second case aa is a vector of 100 elements, so sizeof will return 100*sizeof(char) = 100*1.
In C vector names are also pointers to the first element, but that doesn't mean that aa is actually a pointer.

And by vector, XayC actually means array.

aha
so in arrays it returns the size of elements * size of array
but pointers return the size of first element pointed to by the pointer
thanks

Quote:

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but pointers return the size of first element pointed to by the pointer

---

No, using the sizeof operator on a pointer will return the size of the pointer, no matter what type it points to.

"man sizeof"

returns the size (in number of bytes) of the argument passed in.

pointers = 4 bytes (on most desktop)

char = 1 byte (again on most desktop)

> char = 1 byte (again on most desktop)
Always, actually. (The catch is that in C/C++, a byte is _at least_ 8 bits, not necessarily exactly 8.)

In C at least sizeof char is always 1:

Quote:

---

When applied to an operand that has type char , unsigned char , or signed char , (or a qualified version thereof) the result is 1.

---

http://web.archive.org/web/200502070...t.html#3.3.3.4

thanks all
" When applied to an operand that has type char , unsigned char , or signed char , (or a qualified version thereof) the result is 1. When applied to an operand that has array type, the result is the total number of bytes in the array"
thanks allot vart

Shouldn't

Code:

---

if((aa=malloc(100 *sizeof(char)))==NULL)

---

be:

Code:

---

if((aa=malloc(100 *sizeof(char *)))==NULL)

---

Absolutely not. A char * points to chars, not other char *'s.

To make it even less problematic, you should write something like (100 * sizeof(*aa)) for the size. That way, if you have to change the type of variable aa, you don't have to worry about the memory allocation size.

sorry i was just going through a guide and came across this:

Code:

---

int main(void)
{
    char **p;
    int i;
    p = malloc(sizeof(char*) * 10);  // allocate 10 char*s-worth of bytes
    for(i = 0; i < 10; i++) {
        *(p+i) = malloc(30); // 30 bytes for each pointer
        // alternatively we could have written, above:
        //  p[i] = malloc(30);
        // but we didn't.
        sprintf(*(p+i), "this is string #%d", i);
    }
    for(i = 0; i < 10; i++) {
        printf("%d: %s\n", i, p[i]); // p[i] same as *(p+i)
    }
    return 0;
}

---

Hence, why I suggested it.

And it's correct there, because p is a pointer to a pointer in that case.

that's right you are using array of pointers
that's new for me thanks for adding it :)

Quote:

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And it's correct there, because p is a pointer to a pointer in that case.

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Oh sorry, I completely misread the *aa. I thought s/he put **a. My mistake.