memory allocation
hi
i am try to create a dynamic array of char
this is what i have done
Code:char *temp; unsigned int tempSize = 0; tempSize = 8212; temp = (char *) malloc(tempSize); printf("size of temp %d\n", sizeof(temp));
what gets printed is
size of temp 4
shouldnt the size of temp be 8212?
Nope. The pointer is still a pointer no matter how much memory you allocate to it. The only difference is, instead of pointing to nothing, it's pointing to 8212 bytes of memory. You can still access it like an array, the memory is there. Also you shouldn't, and in this case don't need to, cast malloc.
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ok thanks =D
And you should also free() dynamically allocated memory.
I can't think of too many cases that you would need to cast malloc(). Unless you've using Turbo C or something.Also you shouldn't, and in this case don't need to, cast malloc.
sizeof(a pointer) will return 4 on 32 bit machines and 2 on 16 bit ones. It won't return the number of bytes allocated to that pointer, as you've found out. There is no way of knowing (that I'm aware of) how to figure out how much memory is allocated to a particular pointer. It's best just to keep the size for future reference.
dwk
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