Question

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嗨，我想逐个输入数字，但我得到的输出是......

#include<stdio.h>

int calsum(int x, int y, int z);

void main()
{
    int a,b,c,i, sum;
    for(i = 1; i <= 3; i++)
        printf("\n Enter Number %d\n", i);

    scanf("%d %d %d", &a, &b, &c);
    sum= calsum(a,b,c);
    printf("\nSum=%d", sum);
}

int calsum(int x, int y, int z)
{
    int d;
    d = x + y + z;
    return (d);
}

我希望它一次向我询问输入。请帮我。我是新手。

Answer 1

您可以使用数组

int a[3], i, sum;
for (i = 0; i < 3 ; i++)
{
    printf("\n Enter Number %d\n", i+1);
    scanf("%d", &a[i]);
}
sum = calsum(a[0], a[1], a[2]);

Answer 2

您的for循环内部只有一行代码，
```
printf("\n Enter Number %d\n", i);
```
所以输出正确。
main()必须返回int。
您必须检查scanf()。

你可以使用这样的东西

#include <stdio.h>

int calsum(int x, int y, int z);
int getinteger();

int main()
{
    int a, b, c, z;

    a = getinteger(1);
    b = getinteger(2);
    c = getinteger(3);

    z = calsum(a, b, c);

    printf("\nSum = %d", z);

    return 0;
}

int calsum(int x, int y, int z)
{
    int d;
    d = x + y  +z;
    return d;
}

int getinteger(int index)
{
    int value;
    printf("Enter the %dth number > ", index);
    while (scanf("%d", &value) != 1)
    {
        int chr;
        while (((chr = getchar()) != '\n') && (chr != EOF))
            continue;
        printf("Invalid input -- try again\n");
        printf("Enter the %dth number > ", index);
    }
    return value;
}

没有获得所需的输出。有人能告诉我解决方案吗？

2 个答案: