Question

How is the first argument passed to a function in x86-64 assembly? Give an example of this happening in assembly and the corresponding C code.

What x86-64 register is changed to allocate local variables? Explain briefly with an example.

Answer 1

to answer the following question I would say

the string does not need to be passed you can pass a pointer but the string needs to be in memory somewhere.

by putting the address of that memory into the argument register

Assembly code is :

/////// declare a data segment object containg the string
    .data
str:
    .string "example"

////////code to call fun(char *) with the string:
    .text
    movq $str, %rdi
    call fun

C code is :

char str[] = "example"

void fun(char *str)

.........................................................

example code ;

#include <stdint.h>

int main(void)
{
    uint64_t a = 0x12345679abcdefULL;
    uint64_t b = 0xfdcba987654321ULL;
    uint64_t c = a + b;
    return 0;
}

if you run the above code u come to know that It seems to allocate two 32-bit values on the stack and add then with addl and adcl separately. thia happens with both m32 and m64 bit switch. all though I was on a 64bit system der seemed no need to add any dedicated instruction set for 64bit integers.