「不能收集變量」？

Question

這是什麼意思：「不能收集變量」？我該如何解決它？它使調試器不會報告值。

我想打電話給我的功能

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

我要去改變和重新排列argv。我的變量string是char **string[100][100];，也許這並不理想。該字符串變量會用新的論據來更新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]); 
    } 
} 

但我做錯了什麼，它崩潰。錯誤表示「無法收集變量」和strdup的段落錯誤。

我也試過它編譯，但也成爲一個分段錯誤如下：* string[j] = * strdup(& argv[j]);

GDB說：

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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) 

我應該使用strcpy代替或更改一些聲明？

Answer 1

argv是一個字符串數組（char *argv[]），它不是一個char數組。如果你的字符串被定義爲char string[100][100];，也就是說，它的存儲被定義爲數組，你可以做到以下幾點：

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]); 
    } 
} 

但這是受崩潰，如果argv比的string數量多串，即是100，或argv的任何元素大於99不再那麼你也可以做到動態分配：

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; 
} 

編輯：添加一個測試程序來演示爲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; 
} 

調用此問題：

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