|
#2
自由而无用2021-10-22 19:25
Can I re-describe your question like this ?
whats the difference between the function prototype void initArr(int i[3]) and void initArr(int *i) -------------------------gcc answer------------------- no difference (maybe its a standard ?) -------------------------the reason------------------- gcc generates same code -------------------------how to try------------------- //online parser: https://www.bccn.net/run/  程序代码：#include <stdio.h> #include <stdlib.h> //#define ARG_TYPE #ifdef ARG_TYPE void initArr(int i[3]) #else void initArr(int *i) #endif { printf("sub[i] = %d\n", i[0]); i[0] = 100; printf("sub[i] = %d\n", i[0]); } int main(int argc, char *argv[]) { int i[3] = {10, 20, 30}; initArr(i); printf("main[i] = %d\n", i[0]); system("gcc *.c -o v.out"); #ifdef ARG_TYPE system("objdump -d v.out > a.txt"); system("cat a.txt"); #else system("objdump -d v.out > b.txt"); system("cat b.txt"); #endif system("diff a.txt b.txt"); return 0; } output sample: ...... < 4004ef: 49 c7 c0 20 07 40 00 mov $0x400720,%r8 < 4004f6: 48 c7 c1 b0 06 40 00 mov $0x4006b0,%rcx --- > 4004ef: 49 c7 c0 30 07 40 00 mov $0x400730,%r8 > 4004f6: 48 c7 c1 c0 06 40 00 mov $0x4006c0,%rcx ...... So, finally you will find that nothing is different except the dynamic address |
程序代码：void initArr(int i[3]) {
printf(" %d ", i[0]);// 打印结果10
i[0] = 100;
printf(" %d ", i[0]);// 打印结果100
}
void test2() {
int i[3] = { 10,20,30 };
initArr(i);
printf(" %d ", i[0]);// 打印结果100
}
为什么我这个主调函数test2()中的局部变量i, 传递给initArr,在initArr中修改i的内容,主调函数也会受到影响?
initArr中的I不该是独立栈空间新开辟的局部变量吗?照说不是只有地址传递用指针接收才能达到这效果吗?
学多级指针的时候老师说,传参相当与赋值.被调函数会自己开辟空间存放变量.
从语法角度说,如果传I就是传递地址,那么被调函数定义形参的时候不应该需要一个指针去间接引用吗?
还是说就单纯的把这个情况看成是语法糖?
