c - 我刚刚设置的变量中 printf 期间的段错误

Question

所以当我在以下情况下调用 printf 时，我遇到了段错误。我只是看不出我做错了什么。有任何想法吗？太感谢了。我已经在代码中用注释标记了出现段错误的位置(在第一段代码中)。

...
    char* command_to_run;
    if(is_command_built_in(exec_args, command_to_run)){
        //run built in command
        printf("command_to_run = %s\n", command_to_run); // <--- this is where the problem is
        run_built_in_command(exec_args);
    }
...

int is_command_built_in(char** args, char* matched_command){
    char* built_in_commands[] = {"something", "quit", "hey"};
    int size_of_commands_arr = 3;
    int i;
    //char* command_to_execute;
    for(i = 0; i < size_of_commands_arr; i++){
        int same = strcmp(args[0], built_in_commands[i]);
        if(same == 0){
            //they were the same
            matched_command = built_in_commands[i];
            return 1;
        }
    }
    return 0;
}

Answer 1

您正在按值传递指针 matched_command。因此，调用 is_command_built_in 不会更改它。所以它保留了它的未初始化值。

试试这个:

char* command_to_run;
if(is_command_built_in(exec_args, &command_to_run)){   //  Changed this line.
    //run built in command
    printf("command_to_run = %s\n", command_to_run); // <--- this is where the problem is
    run_built_in_command(exec_args);
}


int is_command_built_in(char** args, char** matched_command){   //  Changed this line.
    char* built_in_commands[] = {"something", "quit", "hey"};
    int size_of_commands_arr = 3;
    int i;
    //char* command_to_execute;
    for(i = 0; i < size_of_commands_arr; i++){
        int same = strcmp(args[0], built_in_commands[i]);
        if(same == 0){
            //they were the same
            *matched_command = built_in_commands[i];   //  And changed this line.
            return 1;
        }
    }
    return 0;
}