what's the difference between the return value of a malloc() and an array name

Question

for example, I have

char* c=(char *)malloc(100*sizeof(char));
printf("0x%x,c);
printf("0x%x,&c);

and this will shows different value for c and &c.

However,if i m having the following:

char d[100];
printf("0x%x,d);
printf("0x%x,&d);

this shows that value of d and &d are the same.

How comes the first code gives me the different result for c and &c?

Read: [What does sizeof(&arr) return?](http://stackoverflow.com/questions/15177420/what-does-sizeofarr-return/15177499#15177499) Explains difference between `d`, `&d` — Grijesh Chauhan, Jul 15 '13 at 17:40

Carl Norum · Accepted Answer · 2013-07-15T17:42:20.180

An array decays into a pointer to its first element in many contexts, including its use in a function call. It doesn't decay when it's the operand of the unary & (address-of) operator. That means d and &d yield the same address in your example, but have different types. char * and char (*)[100], respectively, in your case.

In contrast, c is a pointer. When you take its address with &, you're getting the address of the pointer variable, as opposed to using c directly, which gives you the address it's pointing to. c is a char *, and &c is a char **.

Editorial note: Use %p to print pointer types. %x is for an unsigned int, and unsigned int might be a different size from a pointer. Corrected code might look like:

printf("%p\n", (void *)c);

what's the difference between the return value of a malloc() and an array name

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