Linux 终端中 C 中用户输入的忘记代码

Question

I've written the code, however I cannot remember how to get the user's input.

The command to run the code is ./rle "HHEELLLLO W" but in my code, I don't know how to get the code to read the "HHEELLLLO W" .

If I have a regular printf function, the code will work and will print H1E1L3O 0W0 but I want it so any user's input can be calculated.

I tried using atoi but I think I've done it wrong and I don't know how to fix it.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_RLEN 50

char* encode(char* src)
{
    int rLen;
    char count[MAX_RLEN];
    int len = strlen(src);

    char* dest = (char*)malloc(sizeof(char) * (len * 2 + 1));

    int i, j = 0, k;


    for (i = 0; i < len; i++) {


        dest[j++] = src[i];


        rLen = 1;
        while (i + 1 < len && src[i] == src[i + 1]) {
            rLen++;
            i++;
        }


        sprintf(count, "%d", rLen);


        for (k = 0; *(count + k); k++, j++) {
            dest[j] = count[k];
        }
    }


    dest[j] = '\0';
    return dest;
}


int main(int argc, char ** argv)
{
    char str[] = atoi(argv[1]);
    char* res = encode(str);
    printf("%s", res);
    getchar();
}

When I compile it, I get this error:

rle.c: In function ‘main’:
rle.c:47:18: error: invalid initializer
     char str[] = atoi(argv[1]);
                  ^~~~
rle.c:45:14: error: unused parameter ‘argc’ [-Werror=unused-parameter]
 int main(int argc, char ** argv)
              ^~~~
cc1: all warnings being treated as errors

Answer 1

atoi converts strings (of digits) to integers, which is nothing like what you to do here.

I would recommend doing this one of two ways:

(1) Use the command-line argument directly:

int main(int argc, char ** argv)
{
    char* str = argv[1];
    char* res = encode(str);
    printf("%s\n", res);
}

This works because argv[1] is already a string, as you want. (In fact in this case you don't even need str ; you could just do char* res = encode(argv[1]); .)

In this case you will have the issue that the shell will break the command line up into words as argv[1] , argv[2] , etc., so argv[1] will contain just the first word. You can either use quotes on the command line to force everything into argv[1] , or use the next technique.

(2) Read a line from the user:

int main(int argc, char ** argv)
{
    char str[100];
    printf("type a string:\n");
    fgets(str, sizeof(str), stdin);
    char* res = encode(str);
    printf("%s\n", res);
}

---

In both cases there's also some additional error checking you theoretically ought to do.

In the first case you're assuming the user actually gave you a command-line argument. If the user runs your program without providing an argument, argv[1] will be a null pointer, and your code will probably crash. To prevent that, you could add the test

if(argc <= 1) {
    printf("You didn't type anything!\n");
    exit(1);
}

At the same time you could double-check that there aren't any extra arguments:

if(argc > 2) {
    printf("(warning: extra argument(s) ignored)\n");
}

In the second case, where you prompt the user, there's still the chance that they won't type anything, so you should check the return value from fgets :

if(fgets(str, sizeof(str), stdin) == NULL) {
    printf("You didn't type anything!\n");
    exit(1);
}

As you'll notice if you try this, there's also the issue that fgets leaves the \n in the buffer, which may not be what you want.