Question

我只是想问为什么我的功能read_int不接受fgetc作为int的输入？（我认为是这种情况，因为我总是输入一个整数但它仍然会转到显示错误的int函数。

这是我的代码（来自我的教授的skel代码）：

int main() {
    int the_int;
    double the_double;
    printf("Doing tests of read_int and read_double functions ...\n\n");
    printf("Please try to enter an int: ");
    the_int = read_int();
    printf("read_int succeeded.  It read a value of %d.\n", the_int);
    printf("Please try to enter a double: ");
    the_double = read_double();
    printf("read_double succeeded.  It read a value of %g.\n", the_double);
    return 0;
}

int read_int(void) {
    int f = fgetc(stdin);
    if (fgetc(stdin) != EOF) {
        printf("Sorry that is invalid.");
        exit(1);
    }
    return f;
}

Answer 1

fgetc()用于从流中读取字节，它一次返回一个字符。

要阅读int，请使用scanf("%d", &f)并阅读double，scanf("%lf", &d) f和d定义为{{1}分别和int f;。

还包括double d;并在调用之前定义函数。

以下是修改后的版本：

<stdio.h>

如果您需要使用#include <stdio.h> int read_int(void) { int f; if (scanf("%d", &f) != 1) { printf("Sorry that input is invalid.\n"); exit(1); } return f; } double read_double(void) { double d; if (scanf("%lf", &d) != 1) { printf("Sorry that input is invalid.\n"); exit(1); } return d; } int main(void) { int the_int; double the_double; printf("Doing tests of read_int and read_double functions ...\n\n"); printf("Please try to enter an int: "); the_int = read_int(); printf("read_int succeeded. It read a value of %d.\n", the_int); printf("Please try to enter a double: "); the_double = read_double(); printf("read_double succeeded. It read a value of %g.\n", the_double); return 0; }来读取输入流，则以下是fgetc()和read_int的修改版本，以将这些字节转换为实际数字：

read_double

使用fgetc和stdin指针。错误处理输入

1 个答案: