交換功能和一般功能

Question

我想我的運氣在這個交換功能，但我有問題。

我希望「num1」應該換成num2的值，反之亦然。

任何人都可以把我推向正確的方向嗎？

#include <stdio.h> 


void swap(int *a, int *b) 
{ 
int temp = *a; 
a = *b; 
b = temp; 
printf("Just checking if this badboy gets to the swapfunction.\n"); 
} 


int main() 
{ 
int num1 = 33; 
int num2 = 45; 

swap(&num1, &num2); 

printf("A: %d\n", num1); 
printf("B: %d\n", num2); 

getchar(); 
return 0; 
} 

Answer 1

您需要尊重指針：

int temp = *a; 
*a = *b; 
*b = temp; 

沒有你的編譯器警告你？

Answer 2

你應該知道一些重要的事情是：需要

int temp = *a; // Correct , temp stores value at the address of a 

a = *b; //Incorrect, a is pointer that is used to hold address NOT values 

更正爲：

*a = *b; //Correct, now value at address of a is = value at address of b 

同樣的錯誤，在這裏也：

b = temp;//Incorrect, cause b is pointer used to hold address and NOT values 

進行糾正是

*b = temp;