c - 为什么我的代码返回不正确的值？

Question

我是 C 的新手，我想弄清楚为什么我的代码返回了不正确的值。

int main()
{


    printf("Welcome to my number generator! \n");
    printf("What is the first number in the range? \n");
    int rng1 = scanf("%d", &rng1);
    printf("What is the second number in the range? \n");
    int rng2 = scanf("%d", &rng2);
    printf("What would increment would you like to go up in? \n");
    int inc = scanf("%d", &inc);

    do
        {
        printf("%d\n"rng1);
        rng1 += inc;
        }
    while(rng1 != rng2);
    }
    return 0;
    }

从这段代码中，我希望第一个范围和第二个范围之间的数字列表以一定的数字上升，但我得到的值是 1。我做错了什么？附言我试图“调试”它并发现当我使用时:

if(isalpha(rng1));
    printf("I am a String...")
if(isdigit(rng1))
    printf("I am a Digit")

它返回，“我是一个字符串...”。

谢谢!

Answer 1

您已经通过调用 scanf 将输入值分配给变量 rng1、rng2 和 inc scanf 还返回成功填充的参数列表的项目数。因此，将 scanf 的返回值赋给这些变量是不正确的。只需使用返回值作为输入的数量。您应该读取值 1，因为您只想为每个 scanf 读取一个值。此外，您还可以检查输入的值，以检测输入的值是否有效。

除此之外，我想对您的代码做一些修改。特别是 do{...}while(); 循环可能会由于 != 运算符而无限期运行。请参阅下面代码中的注释。

int main()
{
    /* Declare the variables and do not assign the return value of scanf */
    int rng1, rng2, inc;
    printf("Welcome to my number generator! \n");
    printf("What is the first number in the range? \n");
    /* repeat this check condition for each scanf, exit( EXIT_FAILURE ) requires #include <stdlib.h>*/

    if (1 != scanf("%d", &rng1)) {
        exit( EXIT_FAILURE );
    }
    printf("What is the second number in the range? \n");
    scanf("%d", &rng2);
    printf("What would increment would you like to go up in? \n");
    scanf("%d", &inc);

    do
    {
        printf("%d\n",rng1);
        rng1 += inc;
    }
    /* Use <= instead of != and think about the case for rng1 is 1 rng2 is 5 and inc is 3, can you detect the end of the loop by adding 3 to the starting point 1 ? */
    while(rng1 <= rng2);

    /* Remove the `}` here */
    return 0;
}