Question

所以我试图从命令行解析命令，如：

cd /mnt/cdrom

来

name =“cd”
argc = 2
argv = {“cd”，“/ mnt / cdrom”，NULL}

#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <sys/stat.h>
void parse(char* line, command_t* command) {

    // TODO: Check if string is empty or null
        while (line[i] != ' ' && line[i] != '\0')
     i++;

     if (line[i] == ' ') {

         }

    // TODO: Split the string on whitespace -> should become an array of char
    // TODO: The first string is the command, the length of tokens[1:] is the length of the arguments, and tokens[1:] are the arguments
    // TODO: Create/fill-in the command struct with the data from above
}

就我所做的而言，我不确定如何在没有字符串函数的情况下拆分它。

Answer 1

嗯，它在急速解决方案中，但它的工作原理..当然你需要动态分配内存在parse函数中，例如某种list（对于不同的参数）和某种缓冲区用于当前参数处理。

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

#define MAX_TOKEN  5
#define MAX_LENGTH 64

void copy_str( char* dst, char* src, int size ) {
    int i = 0;
    for( i = 0; i < size; ++i ) {
        dst[i] = src[i];
    }
}

char** parse( char* line, int size, int* argc ) {
    int i = 0, j = 0;
    char** argv = NULL;
    argv = (char**)calloc( MAX_TOKEN, sizeof(char*) );
    for( i = 0; i < MAX_TOKEN; ++i ) {
        argv[i] = (char*)calloc( MAX_LENGTH, sizeof(char) );
    }
    for( i = 0; i < size; ++i ) {
        if( line[i] == ' ' ) {
            copy_str( argv[*argc], line + j, i - j );
            j = i + 1; // length of token
            (*argc)++;
        }
    }
    // copy last
    copy_str( argv[*argc], line + j, i - j );
    (*argc)++;
    return argv;
}

int main( ) {
    int t = 0, i;
    char* s = "cd /mnt/cdrom";
    char** argv = parse( s, 13, &t );
    for( i = 0; i < t; ++i ) {
        printf( "%s\n", argv[i] );
    }
    t = 0;
    s = "ls -l /home/";
    argv = parse( s, 12, &t );
    for( i = 0; i < t; ++i ) {
        printf( "%s\n", argv[i] );
    }
    return 0;
}

解析没有字符串库的命令行命令

1 个答案: