“无法收集变量”是什么意思?我该如何解决它?它使调试器不会报告值。
我正在尝试调用我的函数
write_argument2(argc, * argv, * string[0]);
我将更改并重新排列 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,它崩溃了。该错误显示“无法收集变量”以及 strdup
的段错误。
我还尝试了以下编译但也成为段错误的方法:* 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
来代替还是更改一些声明?
最佳答案
argv
是一个字符串数组 (char *argv[]
),它不是一个 char 数组。如果你的字符串定义为char string[100][100];
,即它的存储定义为数组,你可以执行以下操作:
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]);
}
}
但是,如果 argv
的字符串数量多于 string
的数量(即 100)或 argv
的任何元素,则可能会崩溃> 大于 99。因此您也可以进行动态分配:
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;
}
调用这个问题:
$ ./a.out a bb ccc dddd
0: ./a.out
1: a
2: bb
3: ccc
4: dddd
关于c - "cannot collect variable"?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37017487/