Question

我想从main调用所有三个函数并在主要的两个函数定义中打印结果我无法找到如何做到这一点我已经创建了我需要的所有功能和指针。

#include <math.h> 
#include <stdio.h>
#define _CRT_SECURE_NO_WARNINGS    //to avoid scanf warning or error
int CalculateSum(int num1, int num2);
int CalculateProduct(int num1, int num2);
void CalculateBoth(int num1, int num2, int*sumPtr, int *prodPtr);
int main(void)
{



}
int CalculateSum(int num1, int num2)
{
    int temp1;
    num1 = 1;
    num2 = 2;
    temp1 = num1 + num2;
    //printf("calculatesum value %d ", temp1);
     return (temp1); 

}
int CalculateProduct(int num1, int num2)
{
    num1 = 3;
        num2 = 4;
    int temp2;
    temp2 = num1*num2;
    printf("calc product  %d ", temp2);
        return temp2;

}


void CalculateBoth(int num1, int num2, int*sumPtr, int *prodPtr)
{
    num1 = 5;
    num2 = 6;
    int temp3 = num1 + num2;
    sumPtr = &temp3;
    int temp4 = num1*num2;
    prodPtr = temp4;
    printf("calc sumptr  %d ", temp3);
    printf("Calc both prodptr  %d ", temp4);

}

Answer 1

嗯，显然你需要一些帮助。你的功能并不像你期望的那样。对于初学者，您将num1和num2传递给函数。为了实现这一点，必须在main中声明它们。除了num1和num2之外，您还需要声明变量来保存前两个函数的返回值，比如sum和prod。这些变量也将作为指向CalculateBoth函数的指针传递，以便它们的值在main()中更新并可用。您只需要main()中的简单定义：

/* declare and initialize variables */
int n1 = 0, n2 = 0, sum = 0, prod = 0;

虽然更改函数中变量的值没有任何问题，但是由于您编写了所有函数，因此完全无法将num1和num2传递给函数。看起来您打算将num1和num2的值传递给每个函数，而不是更改函数中的值。例如，您希望在main()中看到：

/* CalculateSum */
n1 = 1, n2 = 2;
sum = CalculateSum (n1, n2);

/* CalculateProduct */
n1 = 3, n2 = 4;
prod = CalculateProduct (n1, n2);

/* CalculateBoth (passing pointer to have value back in main()) */
n1 = 5, n2 = 6;
CalculateBoth (n1, n2, &sum, &prod);

您的任务的剩余部分是在函数中打印值（sum，prod或both），然后再次在main()中为您投保已成功更新函数范围内的值以及main()（函数本身）。下面显示了在函数和main中简单添加相同的print语句。如果您有疑问，请告诉我。

#include <stdio.h>

#define _CRT_SECURE_NO_WARNINGS    // I'll take your word for it...

int CalculateSum (int num1, int num2);
int CalculateProduct (int num1, int num2);
void CalculateBoth (int num1, int num2, int *sumPtr, int *prodPtr);

int main(void)
{
    /* declare and initialize variables */
    int n1 = 0, n2 = 0, sum = 0, prod = 0;

    /* CalculateSum */
    n1 = 1, n2 = 2;
    sum = CalculateSum (n1, n2);
    printf (" calculatesum value %d\n", sum);

    /* CalculateProduct */
    n1 = 3, n2 = 4;
    prod = CalculateProduct (n1, n2);
    printf (" calc product  %d\n", prod);

    /* CalculateBoth (passing pointer to have value back in main()) */
    n1 = 5, n2 = 6;
    CalculateBoth (n1, n2, &sum, &prod);
    printf(" calc both sumPtr   %d\n", sum);
    printf(" calc both prodPtr  %d\n", prod);


    return 0;
}

int CalculateSum (int num1, int num2)
{
    int sum;
    sum = num1 + num2;

    printf ("\n calculatesum value %d\n", sum);

    return sum; 
}

int CalculateProduct (int num1, int num2)
{
    int prod;
    prod = num1 * num2;
    printf ("\n calc product  %d\n", prod);

    return prod;
}

void CalculateBoth (int num1, int num2, int *sumPtr, int *prodPtr)
{
    int sum = num1 + num2;
    *sumPtr = sum;

    int prod = num1 * num2;
    *prodPtr = prod;

    printf("\n calc both sumPtr   %d\n", sum);
    printf(" calc both prodPtr  %d\n", prod);
}

<强>输出

如果仔细查看代码，您会看到在函数和main中都打印了同样的东西：

$ ./bin/fncinmain

 calculatesum value 3
 calculatesum value 3

 calc product  12
 calc product  12

 calc both sumPtr   11
 calc both prodPtr  30
 calc both sumPtr   11
 calc both prodPtr  30

Answer 2

#include <math.h> 
#include <stdio.h>
#include <conio.h>
#define     _CRT_SECURE_NO_WARNINGS     //to avoid scanf warning or error
int CalculateSum(int num1, int num2);
int CalculateProduct(int num1, int num2);
void CalculateBoth(int num1, int num2, int*sumPtr, int *prodPtr);

int main(void) {
    int x = 5, y = 6, prod, sum;
    sum = CalculateSum(x, y);
    prod = CalulateProduct(x, y);
    printf("sum=%d Product=%d", x, y);
    printf("now calulate both at once");


    x = 4;
    y = 3;
    CalculateBoth(x, y, &sum, &prod);
    printf("%d %d", sum, prod);
    getch();
    return 0;

}

int CalculateSum(int num1, int num2) {
    int temp1;
    temp1 = num1 + num2;
    printf("calculatesum value %d ", temp1);
    return (temp1);

}

int CalculateProduct(int num1, int num2) {
    int temp2;
    temp2 = num1*num2;
    printf("calc product  %d ", temp2);
    return temp2;

}

void CalculateBoth(int num1, int num2, int* sumPtr, int *prodPtr) {
    int temp1, temp2;
    temp1 = num1 + num2;
    *sumPtr = temp1;
    temp2 = num1*num2;
    *prodPtr = temp2;
    printf("calc sumptr  %d ", temp1);
    printf("Calc both prodptr  %d ", temp2);

}

Answer 3

如果要返回多个值，请从函数返回一个数组。

int * CalculateBoth(int num1, int num2, int*sumPtr, int *prodPtr)
{
    num1 = 5;
    num2 = 6;
    int arr[2];
    arr[0] = num1 + num2;
    sumPtr = &temp3;
    arr[1] = num1*num2;
    prodPtr = temp4;
    printf("calc sumptr  %d ", temp3);
    printf("Calc both prodptr  %d ", temp4);
    return arr;
}

然后在主

int main()
{
int *p;
p=CalculateBoth(...);
printf("%d %d",p[0],p[1]);
return 0;
}

如何在C中打印我的主要功能？

3 个答案: