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My goal is to read a file and store it's contents in a char array given the offset and number of bytes to be read. I have written a function for doing the same and it works fine.

Now this function is to be called from somewhere else. So I am trying to declare a variable char * data which will hold the contents returned by the above mentioned function. After declaring I tried to allocate it some memory. (I know how much, as I specify the number of bytes to be read). It goes as follows:

char * data;
char * filename = "alphabet.txt";
int data_size = 10;
printf("data size: %d\n", data_size);
data = (char*) malloc (data_size);
printf("Size allocated to data: %d\n",sizeof(data));
return 0;

This code gives the following output:

data size: 10
Size allocated to data: 8

I don't understand this behavior. Can somebody please explain it to me.

Thanks a lot

shahensha

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5
  • 5
    "How does the malloc function work in C?" - nah, rather "How does the sizeof operator work in C?" Commented Sep 26, 2013 at 19:31
  • possible duplicate of Getting the size of a malloc only with the returned pointer Commented Sep 26, 2013 at 19:32
  • 1
    In c you shouldn't cast the return of malloc to a (char *). You can assign the void pointer to it without problem. Commented Sep 26, 2013 at 19:48
  • 2
    @GavinH: It's actually a bad thing to do, not simply unnecessary. Commented Sep 26, 2013 at 19:50
  • @EdS. agreed - updated wording Commented Sep 26, 2013 at 19:52

4 Answers 4

Reset to default
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This has nothing to do with malloc.

sizeof does its thing at compile time, not runtime. sizeof(data) is the same as sizeof(char*). Your program cannot know at compile time how much memory that pointer refers to.

On the other hand, sizeof(some_array) would work as you expect because array sizes are known at compile time.

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3 Comments

Aah! i see, what I am doing here. Thanks for the quick reply. So, I am using malloc correctly, right? I should simply call the read_file function then..right?
@Shahensha "I am using malloc correctly, right?" - not quite -- see this.
@Shahensha Also, printf("%d", sizeof(stuff)); is undefined behavior -- the sizeof operator yields an object of type size_t which is to be printed using the %zu conversion specifier.
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sizeof char *ptr will give the sizeof the pointer address ptr pointing . Its vary depends on the machine.

char *ptr = malloc(10); malloc will allocate the 10 space and ptr pointing
         to the starting address.

Examine below code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char * data;
int data_size = 10;
printf("data size: %d\n", data_size);
data = malloc ((data_size+1)*sizeof(char));
strcpy(data, "datadatada");
printf("string lenght: %d\n",strlen(data));
printf("Size allocated to data: %zd\n",sizeof(data));
return 0;
}
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1 Comment

You have an off-by-one error (no space for the terminating NUL byte) and you're erroneously casting the return value of malloc(). You are also using %d to print a size_t, which is undefined behavior.
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Just these two lines of code

char * data;
printf("Size allocated to data: %d\n",sizeof(data));

would print (in OP's machine):

Size allocated to data: 8

sizeof returns the size of the argument, here data pointer. It is done at compile time.

malloc has nothing to do with sizeof(data).

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Apparently you made some completely unfounded assumptions about the behavior of sizeof. Apparently you assumed that sizeof can be used to query the size of malloc-ed memory block.

It can't be. sizeof has absolutely nothing to do with that. In fact, C library does not provide you with any means to determine the size of malloc-ed memory block. It is simply impossible. If you want to know how much memory you allocated, you have to devise your own way to remember how much memory you requested from malloc. End of a story.

sizeof evaluates to the size of its operand. In this case you supplied a pointer as its operand, so it evaluated to the size of a pointer, which happens to be 8 on your platform.

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