我试图标记用户输入的Shell程序命令。该程序可以运行,但是我认为它不会填充我声明的参数数组。很抱歉,在我学习过程中似乎没有很多信息,但是我不确定为什么它没有填充参数。任何帮助将不胜感激,如果您需要任何其他信息,请随时询问。

#include "HeaderFile.h"
#include <stdio.h>
#include <stdlib.h>
#define token_delimiter " \n\r"

char **shell_read(char *line, char **param){

   line = NULL;
   ssize_t size = 0;
   getline(&line, &size, stdin);
   //printf("%s", line);
   int i = 0;
   char *line_token;
   line_token = strtok(line, token_delimiter);
   printf("%s", line_token);
   for(i=1; line_token!=NULL; i++){
       param[i] = line_token;
       line_token = strtok(NULL, token_delimiter);

   }
   param[0] = NULL;
   return(param);
}

最佳答案

像这样的东西:

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

#define token_delimiter " \n\r"

char **shell_read(void){//Parameters are not necessary. because they are made in this function.
    char *line = NULL;
    size_t size = 0;
    ssize_t len;
    if(-1 == (len = getline(&line, &size, stdin))){
        free(line);
        return NULL;
    }

    char **param = malloc(sizeof(*param) * ((len + 1)/2 + 1));//last +1 for NULL
    int i = 0;
    char *line_token = strtok(line, token_delimiter);

    while(line_token != NULL){
        param[i++] = strdup(line_token);//need allocate and copy
        line_token = strtok(NULL, token_delimiter);
    }
    param[i] = NULL;
    free(line);

    return param;
}

int main(void){
    putchar('>');fflush(stdout);

    char **tokens = shell_read();
    if(tokens){
        for(char **token = tokens; *token; ++token){
            puts(*token);
            free(*token);
        }
        free(tokens);
    }
    return 0;
}

10-07 16:43