scanf（）可以将非零输入转为零

Question

在下面的代码中，main（）中的scanf（）将其中一个输入数字从非零数字转换为零，如while循环中的调试printf（）所示。 我已经在几个编译器上测试了它，但只是为了得到相同的结果。 请告诉我为什么会这样帮助我。 谢谢。

 #include <stdio.h>

unsigned srl (unsigned x, int k)
{
    /* perform shift arithmetically */
    printf("x = %u, (int) x= %d\n", x, (int) x);
    unsigned xsra = (int) x >> k;
    printf("\nxsra before was: %u\n", xsra);
    unsigned test = 0xffffffff;
    test <<= ((sizeof (int) << 3) - k); // get e.g., 0xfff00...
    printf("test after shift is: %x, xsra & test = %x\n", test, xsra & test);
    if (xsra & test == 0) // if xsrl is positve
        return xsra;
    else
        xsra ^= test;    // turn 1s into 0s

    return xsra;
}

int sra (int x, int k) 
{
    /* perform shift logically */
    int xsrl = (unsigned) x >> k;
    unsigned test = 0xffffffff;
    test << ((sizeof (int) << 3) - k + 1); // get e.g., 0xffff00...
    if (xsrl & test == 0) // if xsrl is positve
        return xsrl;
    else 
                            xsrl |= test;

        return xsrl;
}

int main(void)
{
    int a;
    unsigned b;
    unsigned short n;

    puts("Enter an integer and a positive integer (q or negative second number to quit): ");
    while(scanf("%d%u", &a, &b) == 2 && b > 0)
    {
        printf("Enter the number of shifts (between 0 and %d): ", (sizeof (int) << 3) - 1);
        scanf("%d", &n);
        if (n < 0 || n >= ((sizeof (int)) << 3))
        {
            printf("The number of shifts should be between 0 and %d.\n", ((sizeof (int)) << 3) - 1);
            break;
        }
        printf("\nBefore shifting, int a = %d, unsigned b = %u\n", a, b);
        a = sra(a, n);
        b = srl(b, n);
        printf("\nAfter shifting, int a = %d, unsigned b = %u\n", a, b);
        puts("\nEnter an integer and a positive integer (q or negative second number to quit): ");
    }
    puts("Done!");

    return 0;
}

Answer 1

问题是n是unsigned short n ，其大小小于普通int 。 当你调用scanf("%d", &n); ，如果b的内存位置在n之后，它会将值读入n并可能覆盖现有的b值。

您所要做的就是将有问题的行改为：

scanf("%hu", &n);

h是unsigned short int的修饰符，从这里开始 。