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Why is that all the pointers in C have same size? I am on 64 bit arch.

#include<stdio.h> int main(){ printf("int\t%ld\n",sizeof(int*)); printf("char\t%ld\n",sizeof(char*)); printf("void\t%ld\n",sizeof(void*)); printf("float\t%ld\n",sizeof(float*)); } OP : int 8 char 8 void 8 float 8
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    Why wouldn't they have the same size? Commented Jul 17, 2016 at 14:59
  • They are not guaranteed to be same size but generally always are: stackoverflow.com/questions/1241205/… Commented Jul 17, 2016 at 15:01
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    Suppose one of them had a smaller size, that would put a big limit on where things of that type are allowed to exist in memory. Commented Jul 17, 2016 at 15:06
  • @harold: Which is typical on some architectures. There is more to C than x86/ARM. Commented Jul 17, 2016 at 15:51
  • @Olaf it is a consequence, I didn't say that makes it impossible. It sure makes it inconvenient though. Commented Jul 17, 2016 at 15:55

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Because all you need is 64 bits to index a memory address and all the data types you've listed only need to index a memory address (where the given data 'starts' in memory). Note that this is true for C, but in C++ e.g., member function pointers might be more than that.

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Yes they all point to a memory address. The reason they are still of different datatype like int char etc is for C compiler to do Pointer Arithmetic. For example adding a 'char' pointer would increment by one signifying next byte and four for a 'int'.
@Nishant: that's a consequence of the reason, but not the reason. Had it been the reason, for any T and U where sizeof(T) == sizeof(U), we could've said that T* is the same as U* - which is clearly not the case for e.g. different structs. The reason is type safety and the need for dereference operator to have a definite, specific return type. A consequence is that the compiler can now do meaningful pointer arithmetics.

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