每次从stdin读取char的最佳方法是什么？

Question

So basically I am writing a C program that reads a stream of bytes from standard input and treats the bytes as unsigned integers in the range 0 to 255 inclusive. The program counts how often each value in the range 0 to 255 occurs. It also accepts a nonnegative integer as a command line argument. This command line argument gives the number of lines, ​n, ​of output the program should produce. Thus, if​ n​ is 16, the program should print 16 lines of output showing how often the byte values in the range 0 to 15 inclusive occurred.
Each line should begin with the integer value followed by the count, eg
​0 occurred 1014 times
1 occurred 1201 times
and so on.

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.
Here is my code:

#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;
}

Answer 1

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.

That's because:

while (token != "\n") {

is an error. That should be:

while (token != '\n') {

Your compiler should warn about that error. This is what I get when I compile the program using gcc -Wall :

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]

To be more safe, use:

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

Also, you should change the type used for token from char to int . Return type of fgetc() is int . If your platform uses unsigned type for char , you will run into problems with capturing EOF , which is often -1 .

Answer 2

you may use getchar(); -

int c;

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

}

Answer 3

the following is the posted code, with comments..

// <-- 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
}

Answer 4

Generally, what you're trying to do is accomplished as doing the following:

int ch;

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