c - 如何在 C 语言中将函数的返回值存储在变量中

Question

我编写了一个 C 程序如下:-

#include <stdio.h>
#include <conio.h>
#include <string.h>
char getPositions(int randNo, int guessNo);

void main()
{
    char positions[6];
    clrscr();
    positions = getPositions(5242, 5243);
    printf(positions);
    getchar();

}
char getPositions(int randNo, int guessNo)
{
    char outPut[6];
    int randNoArr[4], guessNoArr[4];
    int c, w, p;

    for(int i=4; i>0; i--){
        randNoArr[i] = randNo%10;
        randNo /= 10;

        guessNoArr[i] = guessNo%10;
        guessNo /= 10;
        //If Number Possitioned Right Place Incress variable c
        if(randNoArr[i] == guessNoArr[i]){
            c++;
        }
    }
    for(int j=1; j<=4; j++){
        if(guessNoArr[j] != randNoArr[1] && guessNoArr[j] != randNoArr[2] && guessNoArr[j] && randNoArr[3] || guessNoArr[j] && randNoArr[4]){
            w++;
        }
    }
    if(guessNoArr[1] == randNoArr[2] || guessNoArr[1] == randNoArr[3] || guessNoArr[1] == randNoArr[4]){
        p++;
    }
    if(guessNoArr[2] == randNoArr[1] || guessNoArr[2] == randNoArr[3] || guessNoArr[2] == randNoArr[4]){
        p++;
    }
    if(guessNoArr[3] == randNoArr[1] || guessNoArr[3] == randNoArr[2] || guessNoArr[3] == randNoArr[4]){
        p++;
    }
    if(guessNoArr[4] == randNoArr[1] || guessNoArr[2] == randNoArr[2] || guessNoArr[2] == randNoArr[3]){
        p++;
    }
    sprintf(outPut, "%dC%dW%dP", c,w,p);
    return outPut;
}

我有两个错误:

Error prog1.CPP 10: Lvalue required //positions = getPositions(5242, 5243);
Error prog1.cpp 50: Cannot convert 'char *' to 'char' //return outPut;

Answer 1

首先你犯了一个非常可怕的错误:

尝试将函数中本地声明的数组返回到另一个函数。返回数组的正确做法是返回指向它的指针。并确保指针指向您拥有的内存。当函数返回时，局部变量将失去焦点，因此当函数返回时，您不拥有它们。

为您提供快速解决方案:将位置数组作为第三个参数传递并使函数返回 void 现在对您来说是最简单的事情。