如何在C ++中更改char值的日期？

Question

我是C ++编程的初学者。

主题具有“更改的char值”。 但！ 我的代码总是调用调试错误...。怎么回事.... :(

例

int main(){

    char c[] = "hey";
    char d[] = "hellow";

    cout <<"befor/"<< c << "," << d << endl;
    f(c, d);

    cout <<"after/"<< c << "," << d << endl;


    save();
}
void f(char* p, char* q) {

    int x = *(char*)p;
    int y = *(char*)q;

    int temp= x;
    y= x;
    y = temp;


}

我想获得这个价值

以前，嘿，海洛

在/之后，嘿

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Answer 1

我假设您想交换两个字符串，但这是行不通的，我将尝试解释原因。

int main(){

    char c[] = "hey";
    char d[] = "hellow";

    cout << c << "," << d << endl;
    f(c, d);

    cout << c << "," << d << endl;


    save();
}
void f(char* p, char* q) { 
// When you pass the two strings to this function the arrays turn into
// pointers to the string. 

    int x = *(char*)p; // You now say you want x to be equal to the first 
                       // letter of what p points to, which is "hey". 
                       // So now x equals 'h', although its value is as 
                       // an int.
    int y = *(char*)q; // Here you assign the first letter of "hellow" 
                       // to y. y now equals 'h' also

    int temp= x; // temp now equals 'h'
    y = x; // y now equals 'h' also
    y = temp; // y now equals 'h' also

    // Basically this whole thing does nothing.

}

通过执行此操作，可以使函数交换指针，尽管可能有太多内容无法解释。

#include <iostream>

using namespace std;

void f(char** p, char** q);

int main() {

    char* c = "hey";
    char* d = "hellow";

    cout << "befor/" << c << "," << d << endl;

    f(&c, &d);

    cout << "after/" << c << "," << d << endl;

}
void f(char** p, char** q) {

    char* temp = *p;

    *p = *q;
    *q = temp;
}

为什么必须将函数参数用作指向指针的指针或指向指针的引用，因为仅指针会给函数一个副本，而副本不足以更改原始指针。 抱歉，如果所有这些都没有道理，则您的代码有很多错误，我无法真正解释所有内容。