每次从 stdin 读取 char 的最佳方法是什么?

What is the best way to read a char each time from stdin?

所以基本上我正在编写一个 C 程序,它从标准输入读取字节流并将字节视为 0 到 255 范围内的无符号整数。该程序计算 0 到 255 范围内的每个值出现的频率。它还接受一个非负整数作为命令行 争论。此命令行参数给出程序应生成的输出行数 n。因此,如果 n 为 16,程序应打印 16 行输出,显示 0 到 15 范围内的字节值出现的频率。
每行应以整数值开头,后跟计数,例如
0出现1014次
1 出现 1201 次
等等。

我尝试每次从标准输入读取一个字符并检查它是否为“\n”。然而,条件 (token != "\n") 永远不会 returns False 并且循环永远不会中断。
这是我的代码:

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

int main(int argc, const char* argv[]) {
    char token;
    int n;
    if (argc != 2) {
        printf("Error!\n");
        exit(0);
    }

    n = atoi(argv[1]);
    int i;
    int freq[n];

    for(i = 0; i<n; i++) {
        freq[i] = 0;
    }

    int value;
    printf(">");
    token = fgetc(stdin);

    while (token != "\n") {
        printf("here!");
        value = token;
        if (value < n) {
            freq[value] ++;
        }
        token = fgetc(stdin);
    }
    printf("there");

    for(i = 0; i<n; i++) {
        printf("%d  occured %d times\n",i, freq[i]);
    }

    return 1;
}

您可以使用 getchar(); -

int c;

while((c=getchar())!=EOF){

}

I try to read a char each time from stdin and check if it's "\n". However the condition (token != "\n") never returns False and the loop is never broken.

那是因为:

while (token != "\n") {

是一个错误。那应该是:

while (token != '\n') {

您的编译器应该对该错误发出警告。这是我使用 gcc -Wall:

编译程序时得到的结果
soc.c: In function ‘main’:
soc.c:25:18: warning: comparison between pointer and integer [enabled by default]
     while (token != "\n") {
                  ^
soc.c:25:18: warning: comparison with string literal results in unspecified behavior [-Waddress]

为了更安全,使用:

while (token != '\n' && token != EOF ) {

此外,您应该将用于 token 的类型从 char 更改为 int。 Return fgetc() 的类型是 int。如果您的平台对 char 使用 unsigned 类型,您将 运行 遇到捕获 EOF 的问题,通常是 -1.

下面是贴出的代码,有注释..

// <-- strongly suggest reading the manual for the system functions you use
// so the error on the returned value from fgetc() would not have happened

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

int main(int argc, const char* argv[])
{
    char token;   // <-- fgetc and getchar return integer, not character
    int n;

    if (argc != 2)
    {
        printf("Error!\n"); // <-- this should be a 'usage' statement
        exit(0);            // <-- this exit is due to an error, returned value should be a non-zero value to indicate an error occurred
    }

    // implied else, right number of parameters.

    n = atoi(argv[1]);  // <-- check that 'n' is not zero and not negative and <= 255
                        // <-- output an error message and exit of value is not good

    // implied else, parameter valid

    int i;
    int freq[n];

    for(i = 0; i<n; i++)
    {
        freq[i] = 0;
    }

    int value;
    printf(">");


    // <-- suggest modification to while(), so fgetc() embedded in while statement
    token = fgetc(stdin);
    while (token != "\n") { // <-- this is a comparison between pointer and integer, compiler warning
                            // <-- comparions with string literal results in unspecified behavior
        printf("here!");
        // <-- token, after correcting declartion to int eliminates the need for 'value'
        value = token;
        if (value < n) {
            freq[value] ++;
        }
        token = fgetc(stdin);
    }
    printf("there");

    for(i = 0; i<n; i++) {
        printf("%d  occured %d times\n",i, freq[i]);
    }

    return 1;  // <-- this is the 'good' exit, so should return 0
}

通常,您要执行的操作是通过执行以下操作来完成的:

int ch;

while((ch = getchar()) != '\n' && ch != EOF)
{
   // ...
}