Question

在下面的示例中，如果我在Mac OS X终端中输入一个字符，程序将陷入无限循环，逐行打印Please enter a number:并且永远不允许用户输入任何内容。这段代码出了什么问题？修复是什么？我想以一种方式更改代码，如果未输入数字，将提示用户输入错误消息并要求再次输入数字。

#include <stdio.h>

int main(int argc, const char * argv[]) {

    int number = 0, isnumber;
    getagin: printf("Please enter a number:\n");
    isnumber = scanf("%i", &number);
    if(isnumber) {
        printf("You enterd a number and it was %i\n", number);
    } else {
        printf("You did not eneter a number.\n");
        goto getagin;
    }

    return 0;
}

编辑：我在阅读建议后编辑了代码，并修复了无限循环问题。这对于无限循环问题来说并不是一个糟糕的修复，并且通过一个简单的for循环，我告诉C搜索任何无数字字符。以下代码不允许输入123abc。

#include <stdio.h>
#include <ctype.h>
#include <string.h>

int main(int argc, const char * argv[]) {

    char line[10];
    int loop, arrayLength, number, nan;
    arrayLength = sizeof(line) / sizeof(char);
    do {
        nan = 0;
        printf("Please enter a number:\n");
        fgets(line, arrayLength, stdin);
        for(loop = 0; loop < arrayLength; loop++) { // search for any none numeric charcter inisde the line array
            if(line[loop] == '\n') { // stop the search if there is a carrage return
                break;
            }
            if((line[0] == '-' || line[0] == '+') && loop == 0) {
                continue;
            } // Exculude the sign charcters infront of numbers so the program can accept both negative and positive numbers
            if(!isdigit(line[loop])) { // if there is a none numeric character then add one to nan and break the loop
                nan++;
                break;
            }
        }
    } while(nan || strlen(line) == 1); // check if there is any NaN or the user has just hit enter
    sscanf(line, "%d", &number);
    printf("You enterd number %d\n", number);
    return 0;
}

Answer 1

输入非数字时无限循环的原因是缓冲区中剩余的非数字字符，因为scanf无法读取scanf的下一个读数（因为它与格式说明符不匹配）。在下一次迭代scanf再次找到此字符并且不读取它并立即退出。这种情况反复发生，您将获得无限循环。放置while(getchar() != '\n');语句以使用此字符。

试试这个

#include <stdio.h>

int main(int argc, const char * argv[]) {

    int number = 0, isnumber;
    getagin: printf("Please enter a number:\n");
    isnumber = scanf("%i", &number);
    if(isnumber) {
        printf("You enterd a number and it was %i\n", number);
    } else {
        printf("You did not eneter a number.\n");
        while(getchar() != '\n'); // T consume all non-digits
        goto getagin;
    }

    return 0;
}

为什么这个例子卡在C的无限循环中？

1 个答案: