why using 8 byte register to carry 8 byte type but not using 4 byte register for 4 byte type

This is the code snippet that i am looking

#include <cstdint>
int main() {

    unsigned int i = 0x11111111;
    unsigned int j = 0xFFFFFFFF;

    uint64_t k = 0xFFFFFFFF11111111;
0000000140001000  sub         rsp,18h  
0000000140001004  mov         dword ptr [rsp],11111111h  
000000014000100B  mov         dword ptr [rsp+4],0FFFFFFFFh  
0000000140001013  mov         rax,0FFFFFFFF11111111h  
000000014000101D  mov         qword ptr [rsp+8],rax  
0000000140001022  xor         eax,eax  
0000000140001024  add         rsp,18h  
0000000140001028  ret

We are using rax register to carry uint64_t value.

Why we are not using eax or some other 32 bit register for int?

>Solution :

Since you’re not compiling with optimization, it is not using registers to hold any of your variables — they’re all in the stack frame (at rsp, rsp+4,and rsp+8)

rax here is just used here to temporarily hold the constant 0xFFFFFFFF11111111 as there’s no single instruction to write a 64 bit value to memory. There is an instruction to write a 32-bit value to memory, so that is used for the unsigned int initializations

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