Question

因此，当我在以下情况下调用printf时，我会发生错误。我只是看不出我做错了什么。有任何想法吗？太感谢了。我在代码中标记了一个地方，在那里我得到了一个带有注释的seg错误（在第一个代码块中）。

...
    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

您按值传递指针matched_command。因此，is_command_built_in的调用不会改变它。所以它保留了它未初始化的价值。

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;
}

Answer 2

command_to_run未初始化。对is_command_built_in的调用可能很容易崩溃。这就是未定义行为的本质。

我刚设置的变量printf期间的段错误

2 个答案: