c - 如何将 char ** 转换为 c 中的 char *[]？

Question

execv 函数将指针数组作为第二个参数。我有一个指向指针的指针，一个动态创建的字符串列表。

我如何从中创建一个指针数组？

char **list = malloc((argc)*sizeof(char*));
int i=0;
for(i=0;i<argc;++i){ // using argv for example...
 list[i] = malloc(strlen(argv[i])+1);
 strcpy(list[i], argv[i]);
}
// create from list an array of pointers
execv(list_pointers[0], list_pointers);

否则，如果将 list 简单地传递给 execv，我会得到一个 Bad address 错误。

Answer 1

在 header unistd.h 中声明的函数execv 的签名是

int execv(const char *path, char *const argv[]);

请注意，这与

相同

int execv(const char *path, char *const *argv);

这意味着 argv 是一个指向 char * const 类型对象的指针，即一个指向字符的常量指针。此外，execv 的手册页说 -

The first argument, by convention, should point to the filename associated with the file being executed. The array of pointers must be terminated by a NULL pointer.

此外，类型为char ** 的list 赋值与execv 的第二个参数兼容。我建议进行以下更改 -

// +1 for the terminating NULL pointer required for the 
// second argument of execv

char **list = malloc((argc + 1) * sizeof *list); 
if(list == NULL) {
    printf("not enough memory to allocate\n");
    // handle it
}
int i = 0;
for(i = 0; i < argc; ++i) {
    // strdup returns a pointer to a new string
    // which is a duplicate of the string argv[i]
    // this does effectively the same as the commented 
    // out block after the below statement.
    // include the header string.h for the prototype
    // of strdup POSIX function.

    list[i] = strdup(argv[i]);

    /* 
    list[i] = malloc(strlen(argv[i])+1);
    if(list[i] == NULL) {
        printf("not enough memory to allocate\n");
        // handle it
    }
    strcpy(list[i], argv[i]);
    */
}

list[argc] = NULL;  // terminate the array with the NULL pointer
execv(list[0], list);