C 语言：收到错误：警告：格式“%d”需要“int *”类型的参数，但参数 2 的类型为“int **”[-Wformat=]

Posted 2023-02-21

【中文标题】C 语言：收到错误：警告：格式“%d”需要“int *”类型的参数，但参数 2 的类型为“int **”[-Wformat=]

【英文标题】：C language: received error: warning: format ‘%d’ expects argument of type ‘int *’, but argument 2 has type ‘int **’ [-Wformat=]

【发布时间】：2021-11-14 11:54:46

【问题描述】：

仅限 C 语言。

当我运行此代码时，我在第 7 行收到错误消息：

warning: format ‘%d’ expects argument of type ‘int *’, but argument 2 has type ‘int **’ [-Wformat=]

另外，我在第 8 行收到错误消息：

warining: function returns address of local variable [-Wreturn-local-addr]

我附上了一张我想要输出的图片。有人可以显示更正的代码吗，不胜感激，谢谢！

#include <stdio.h>

// Method getFraction() asks for numerator and denominator
int *getFraction() 
    int *n[2];  // declare array of pointers to return the numerator and
                // denominator
    printf("\n Enter the numerator and the denominator -- ");
    scanf("%d%d", (n + 0), (n + 1));  // read numerator and denominator
    return n;                         // return n


// method smallest() takes twointegers as its parameter and return the smallest
int smallest(int a, int b) 
    if (a < b)     // condition for a is smallest
        return a;  // return a
    else
        return b;  // return b


// recursive method to return the GCD() of two numbers
int gcd(int a, int b) 
    if (b == 0)  // if denominator is 0 then return a
        return a;
    // recursively call to gcd()
    return gcd(b, a % b);


// method reduce() takes two parameters as call by reference and
// reduce the fraction
void reduce(int *a, int *b) 
    int g;
    // call to gcd()
    g = gcd(*a, *b);
    // reduce numerator
    *a = *a / g;
    // reduce denominator
    *b = *b / g;


// driver code
int main() 
    int *num, a, b, small, g;
    char ch;
    // loop to repeat the process till user wants
    do 
        num = getFraction();  // call to getFraction();
        a = *(num + 0);  // assign numerator to a
        b = *(num + 1);  // assign denominator to b
        // condition for denominator not zero
        if (b != 0) 
            reduce(&a, &b);  // call to reduce() as call by reference
            // print the reduced fraction
            printf("\n The reduced fraction is -- %d / %d", a, b);
        
        else printf("\n Denominator should not be zero.");
        // ask user to repeat more
        printf("\n Try again (Y/N) -- ");
        fflush(stdin);
        scanf("%c", &ch);  // read users choice
     while (ch == 'y' || ch == 'Y');
    return 0;

【问题讨论】：

为什么是int *n[2]; 而不是int n[2];？你永远不会分配内存，所以读入n[0] 和n[1] 会使程序有未定义的行为

请正确格式化您的代码，以便可以按原样复制并成功编译。目前，它的格式很差。

你传递n 这是数组本身的地址（不是它未初始化的指针内容）。建议你重构一下函数，比如void getFraction(int n[2])

【参考方案1】：

#include <stdio.h>

// Method getFraction() asks for numerator and denominator
int *getFraction() 
    int n[2];// declare array of pointers to return the numerator and
                // denominator
    int *n1=n;
    printf("\n Enter the numerator and the denominator -- ");
    scanf("%d%d", &n[0], &n[1]);  // read numerator and denominator
    return n1;                         // return n


// method smallest() takes twointegers as its parameter and return the smallest
int smallest(int a, int b) 
    if (a < b)     // condition for a is smallest
        return a;  // return a
    else
        return b;  // return b


// recursive method to return the GCD() of two numbers
int gcd(int a, int b) 
    if (b == 0)  // if denominator is 0 then return a
        return a;
    // recursively call to gcd()
    return gcd(b, a % b);


// method reduce() takes two parameters as call by reference and
// reduce the fraction
void reduce(int *a, int *b) 
    int g;
    // call to gcd()
    g = gcd(*a, *b);
    // reduce numerator
    *a = *a / g;
    // reduce denominator
    *b = *b / g;


// driver code
int main() 
    int *num, a, b, small, g;
    char ch;
    // loop to repeat the process till user wants
    do 
        num = getFraction();  // call to getFraction();
        a = *(num + 0);  // assign numerator to a
        b = *(num + 1);  // assign denominator to b

        // condition for denominator not zero
        if (b != 0) 
            reduce(&a, &b);  // call to reduce() as call by reference
            // print the reduced fraction
            printf("\n The reduced fraction is -- %d / %d", a, b);
        
        else printf("\n Denominator should not be zero.");
        // ask user to repeat more
        printf("\n Try again (Y/N) -- ");
        fflush(stdin);
        scanf("%c", &ch);  // read users choice
     while (ch == 'y' || ch == 'Y');
    return 0;

【讨论】：

成功了！谢谢！

@yessirblender 由于这个答案有帮助，请通过单击灰色复选标记来考虑accepting。

【参考方案2】：

你在第 8 行得到一个错误，因为在你的函数返回它的值堆栈弹出局部变量之后，我们无法访问它们的地址，因此你返回的东西可能会导致你的操作系统被杀死。编译器在构建时出错。