malloc分配的内存大小

Question

I am assigning a new memory chunk to a pointer, but apparently the size of the chunk is not the one which I pass as a parameter to malloc

char *q="tre";
printf("q in main %zu\n", sizeof(q));
q = (char*)malloc(6);
printf("q in main %zu\n", sizeof(q));

Outputs

8
8

The pointer however does point to a new memory chunk. How is this possible?

Answer 1

sizeof returns size of pointer, in your case it is (char*) , it will not give the memory allocated by the malloc . Keep the memory size in separate variable for later use.

Answer 2

char *q;
printf("%zu\n", sizeof(q));

sizeof(q) refers to the size of the pointer, not the amount of memory it points to.

Answer 3

What you are obtaining is the size of the variable q as a pointer type. In general all pointers will have the same size in your program.

Since 8 bytes are 64 bits, it seems you are doing 64-bit applications. :)

Answer 4

sizeof(q) returns the size of the pointer q which will on a 64 bit machine be 8 bytes, not the size of the memory block allocated at that pointer. sizeof is a compile time not a runtime operation.

I'm not clear what you want to do here, but if you want to allocate enough memory for a string at location s , then you want to malloc(strlen(s)+1) (+1 for the terminating NULL ).

Perhaps you want to get the size of malloc() ed block. There is not a portable way to do this to my knowledge, but malloc_usable_size nearly does it on glibc . From the man page:

malloc_usable_size() returns the number of bytes available in the dynamically allocated buffer ptr, which may be greater than the requested size (but is guaranteed to be at least as large, if the request was successful). Typically, you should store the requested allocation size rather than use this function.

Note the last sentence.