Question

我试图用C编写一个简单的程序，该程序获取输入的名称，小时费率和小时数，并计算总薪水，税金，净薪水等，但是由于我在制作一个可以打印出结果的函数而陷入困境。

我创建了两个循环：一个循环产生加班，总工资，税金和净工资。打印输出生成的数据的另一个循环。第一个工作正常。它生成的数据很好，但是在我运行程序时，它只是忽略了第二个循环并完成了程序。

这只是一个简单的void函数，没有返回值，但是我被卡住了。我通过引用传递。如果您知道怎么了，请告诉我。

#include<stdio.h>
#include<stdlib.h>
#include <string.h>
#define size 5

char name[size][20];
float rate[size];
float hours[size];
int i = 0;

void input();
float getbasepay(float *rt, float *hr);
float getovtime(float *rt, float *hr);
float getgross(float *base, float *ovt);
float gettax(float *gross);
float getnetpay(float *gross, float *tax);
void printout(char name, float rate, float hours, float base, float  ovt, float gross, float tax, float net);


void input()
{   
    for (i = 0; i < size; i++)
    {
        puts("\ntype name: (type -1 to quit) \n");
        scanf_s("%19s", &name[i], 20);
        if (strcmp(name[i], "-1") == 0) { break; }
        puts("\ntype hourly rate: (type -1 to quit) \n");
        scanf_s("%f", &rate[i]);
        if (rate[i] == -1) { break; }
        puts("\ntype hours worked: (type -1 to quit) \n");
        scanf_s("%f", &hours[i]);
        if (hours[i] == -1) { break; }
    }
    return ;
}

float getbasepay(float *rt, float *hr)
{   float base;
    float baseo;
    float excessive = (*hr - 40);

    if (*hr <= 40) 
    {
        base = (*rt) * (*hr);
        return base;
    }
    if (*hr >= 40)
    {
        baseo = (*hr - excessive) * (*rt);
        return baseo;
    }
}

float getovtime(float *rt,float *hr)
{
    float time;
    float ovt;
    if (*hr >= 40) 
    {
        time = (*hr - 40);
        ovt = time * (*rt) *1.5;

        return ovt;
    }
    else  
    {
        return 0;
    }

}

float getgross(float *base, float *ovt)
{
    float gross = *base + *ovt;
    return gross;
}

float gettax(float *gross)
{
    float tax = *gross * 0.20;

    return tax;
}

float getnetpay(float *gross, float *tax)
{
    float net;
    net = *gross - *tax;
    return net;
}

void printout(char *name, float *rate, float *hours, float *base, float * ovt, float *gross, float *tax, float *net)
{
    printf("\nPay to: %s \n", *name);
    printf("Hourly rate: %f \n", *rate);
    printf("Hours: %f \n", *hours);
    printf("Base pay: $%f \n", *base);
    printf("Overtime: $%f \n", *ovt);
    printf("Gross Pay: $%f \n", *gross);
    printf("Tax:  $%f \n", *tax);
    printf("Net Pay: $%f \n", *net);
}

int main(void)
{
    int o, z;
    float overtime[size] = { 0,0,0,0,0, };
    float basepay[size] = { 0,0,0,0,0, };
    float tax[size] = { 0,0,0,0,0, };
    float grosspay[size] = { 0,0,0,0,0, };
    float netpay[size] = { 0,0,0,0,0, };
    float sum;
    input();

    for (o = 0; o < i; o++) 
    {       
        basepay[o] = getbasepay(&rate[o], &hours[o]);
        overtime[o] = getovtime(&rate[o], &hours[o]);
        grosspay[o] = getgross(&basepay[o], &overtime[o]);
        tax[o] = gettax(&grosspay[o]);
        netpay[o] = getnetpay(&grosspay[o], &tax[o]);
    }

    for (z = 0; z > i; z++)
    {
        printout(&name[z], &rate[z], &hours[z], &basepay[z], &overtime[z], &grosspay[z], &tax[z], &netpay[z]);
    }
}

Answer 1

您的代码无法编译：

gcc -o main main.c

答案：

error: conflicting types for ‘printout’

确实，声明中的签名与您的签名不同写在实现中（例如char * name代替char name等）

Printf函数通过引用传递

1 个答案: