＆＃34;无法收集变量＆＃34;？

Question

这是什么意思：＆＃34;无法收集变量＆＃34; ？我该如何解决？它使调试器不会报告值。

我试图调用我的功能

write_argument2(argc, * argv, * string[0]);

我要改变并重新排列argv。我的变量string是char **string[100][100];，也许这并不理想。字符串变量将使用新参数更新argv：

void write_argument2(int argc, char argv[], char *string[]) {
    int j = 0;
    for (j = 0; j <  argc; j++) {
        if (argv[j])
             string[j] = strdup(&argv[j]);
    }
}

但是我做错了什么而且崩溃了。错误说＆＃34;无法收集变量＆＃34;以及strdup的分段错误。

我还尝试了以下编译，但也成为分段错误：* string[j] = * strdup( & argv[j]);

Gdb说：

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Reading symbols from ./shell...done.
(gdb) run
Starting program: /home/dac/ClionProjects/shell2/openshell/shell 
'PATH' is set to /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games:/usr/local/games:/snap/bin.
$ ls

Program received signal SIGSEGV, Segmentation fault.
0x0000000000403b1a in write_argument2 (string=<optimized out>, argv=<optimized out>, 
    argc=<optimized out>) at main.c:147
147              string[j] = strdup( &argv[j]);
(gdb) 

我应该使用strcpy代替还是更改某些声明？

Answer 1

argv是一个字符串数组（char *argv[]），它不是char数组。如果您的字符串定义为char string[100][100];，即其存储被定义为数组，则可以执行以下操作：

void write_argument2(int argc, char *argv[], char string[][100]) {
    int j = 0;
    for (j = 0; j <  argc; j++) {
        if (argv[j])
             strcpy(string[j], argv[j]);
    }
}

但如果argv的字符串多于string的字符串，即100，或argv的任何元素长于99，则可能会崩溃。所以你可以也做动态分配：

char **write_argument2(int argc, char *argv[]) {
    char **string = malloc(argc * sizeof(char*));
    int j = 0;
    for (j = 0; j <  argc; j++) {
        if (argv[j])
             string[j] = strdup(argv[j]);
    }
    return string;
}

编辑：添加一个测试程序来演示OP：

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

char **write_argument2(int argc, char *argv[]) {
    char **string = malloc(argc * sizeof(char*));
    int j = 0;
    for (j = 0; j <  argc; j++) {
        if (argv[j])
             string[j] = strdup(argv[j]);
    }
    return string;
}


int main(int argc, char *argv[])
{
    int i;
    char **string = write_argument2(argc, argv);
    for (i = 0; i < argc; i++) {
        printf("%d: %s\n", i, string[i]);
    }
    return 0;
}

调用此问题：

$ ./a.out a bb ccc dddd
0: ./a.out
1: a
2: bb
3: ccc
4: dddd