1

您好，我正在练习我的 C 语言知识我正在尝试制作一个简单的计算器，但我遇到了这个警告Implicit Declaration of Function，但我调用的函数已被执行。我试图用这个修复它，void start();但该功能没有执行。

成功执行了该功能start();，但有一个隐含的警告：

#include <stdio.h>
#include <conio.h>
#include <stdlib.h>

void addition()
{
    int vala, valb, resu;
    system("cls");
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("ADDITION\n");
    printf("\n");
    printf("Enter the first value of addend: ");
    scanf("%d", &vala);
    printf("Enter the second value of addend: ");
    scanf("%d", &valb);
    resu=vala+valb;
    printf("The sum of %d and %d is: %d\n", vala, valb, resu);
    printf("PRESS [ANY KEY] TO CONTINUE...");
    getch();

    start(); \\THIS CODE
}

void start()
{
    char ope;
    system("cls");
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("What operation will be used:");
    scanf("%s", &ope);

    if (ope == 'a')
    {
        addition();
    }
    else if (ope == 'b')
    {
        printf("bbbbbbbbbbbb\n");
    }
    else
    {
        printf("ccccccccccc\n");
    }

}

int main()
{
    int choices;
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("choose an option:");
    scanf("%d", &choices);

    if (choices == 1)
    {
        start();
    }

    getch();
    return 0;
}

无法执行函数void start();启动但没有隐式警告：

#include <stdio.h>
#include <conio.h>
#include <stdlib.h>

void addition()
{
    int vala, valb, resu;
    system("cls");
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("ADDITION\n");
    printf("\n");
    printf("Enter the first value of addend: ");
    scanf("%d", &vala);
    printf("Enter the second value of addend: ");
    scanf("%d", &valb);
    resu=vala+valb;
    printf("The sum of %d and %d is: %d\n", vala, valb, resu);
    printf("PRESS [ANY KEY] TO CONTINUE...");
    getch();

    void start(); \\THIS CODE
}

void start()
{
    char ope;
    system("cls");
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("What operation will be used:");
    scanf("%s", &ope);

    if (ope == 'a')
    {
        addition();
    }
    else if (ope == 'b')
    {
        printf("bbbbbbbbbbbb\n");
    }
    else
    {
        printf("ccccccccccc\n");
    }

}

int main()
{
    int choices;
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("choose an option:");
    scanf("%d", &choices);

    if (choices == 1)
    {
        start();
    }

    getch();
    return 0;
}
4

2 回答 2

1

start()是在之后声明addition()，但是在addition()你调用start()中，所以编译器不知道是什么start()。此外，在start()你也调用addition()，所以最好的解决方法是使用前向声明：

#include <stdio.h>
#include <conio.h>
#include <stdlib.h>

void start(void); /* forward declaration */
void addition(void); /* forward declaration */

void addition(void)
{
    int vala, valb, resu;
    system("cls");
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("ADDITION\n");
    printf("\n");
    printf("Enter the first value of addend: ");
    scanf("%d", &vala);
    printf("Enter the second value of addend: ");
    scanf("%d", &valb);
    resu=vala+valb;
    printf("The sum of %d and %d is: %d\n", vala, valb, resu);
    printf("PRESS [ANY KEY] TO CONTINUE...");
    getch();

    start(); 
}

void start(void)
{
    char ope;
    system("cls");
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("What operation will be used:");
    scanf("%s", &ope);

    if (ope == 'a')
    {
        addition();
    }
    else if (ope == 'b')
    {
        printf("bbbbbbbbbbbb\n");
    }
    else
    {
        printf("ccccccccccc\n");
    }

}
于 2021-11-16T06:44:45.120 回答
1

不建议采取允许您调用startfrom的措施，这会addition导致潜在的无限递归，只是为了伪造一个循环。最好挂断start();电话addition并将其替换scanf("%s", &ope);为while (scanf(" %c", &ope) > 0).

于 2021-11-17T17:06:24.490 回答