如何将**argv复制到一个新变量?

Nik*_*ntz -1 c memcpy

我想制作一个变量的副本,比如char **argv使用像memcpy 这样的东西但是怎么样?我可以知道尺寸以及如何参数化它,因为memcpy不会将二星作为参数.我必须写一个循环吗?

我必须在执行之前建立管道.因此,我希望像argv循环的每次迭代一样"生成"变量.

    token = strtok(cmd, "|");
    i = 0;
    while (token != NULL)
    {   
        printf("token %s\n", token);       
        makeArgs(token, &argc, &argv);
        for (int b = 0; b < argc; b++) {
           printf("makeargs %d: %s\n", b, argv[b]);   
        }   

        // Will copy argc characters from array1 to array2
       /* memcpy(makearg,argc, argc*18*sizeof(int));*/    
        shellcommand[i].argv = argv;
        i++;
        token = strtok(NULL, "|");
    }           
   /* do stuff */
   fork_pipes(argc, shellcommand);
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我的目标是建立如下的管道.

/*  who | awk '{print $1}' | sort | uniq -c | sort -n */
static char *cmd0[] = { "who",                0 };
static char *cmd1[] = { "awk",  "{print $1}", 0 };
static char *cmd2[] = { "sort",               0 };
static char *cmd3[] = { "uniq", "-c",         0 };
static char *cmd4[] = { "sort", "-n",         0 };

static char **cmds[] = { cmd0, cmd1, cmd2, cmd3, cmd4 };
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mil*_*bug 5

此代码包含以下内容的深层副本argv:

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

int main(int argc, char** argv)
{
    // allocate memory and copy strings
    char** new_argv = malloc((argc+1) * sizeof *new_argv);
    for(int i = 0; i < argc; ++i)
    {
        size_t length = strlen(argv[i])+1;
        new_argv[i] = malloc(length);
        memcpy(new_argv[i], argv[i], length);
    }
    new_argv[argc] = NULL;

    // do operations on new_argv
    for(int i = 0; i < argc; ++i)
    {
        printf("%s\n", new_argv[i]);
    }

    // free memory
    for(int i = 0; i < argc; ++i)
    {
        free(new_argv[i]);
    }
    free(new_argv);
}
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  • 好的,给他解决方案然后:) (3认同)