Problem with dynamic array in C?

[Jack Drowder]

So I was testing out dynamic arrays in C, and I created this incredibly simple program:

#include <stdio.h>

int main(void) {
    char *cTest = malloc(1);
    printf("%d", sizeof(cTest));
    return 0;
}

The output is:

4

Why doesn't it print 1? Am I missing something?

Thank You!

[ahmed hamdy]

you did not allocate any space

you must declare like that:

datatype* variable =(datatype*)mallco(size*sizeof(datatype));

#include<stdio.h> #include<stdlib.h> int main() { char *var; var=(char*)mallco(1*sizeof(char)); return 0; }

[ahmed hamdy]

your output may be rabbitch from memory

i had test it and output was 2

[Ivan Jozsef Balazs]

You printed the size of a pointer (happening to be pointing to character - but this is irrelevant).
Obviously the size does not depend on the contents of the memory area being pointed to. The size's equaling 4 should not be a big surprise on a 32 bit platform.

[Ivan Jozsef Balazs]

This would print 1
printf("%d", sizeof(*cTest));