结构数组复制如何在 C 中工作？

Question

这是我想要在 C 中运行的代码。

#include<stdio.h>
#include<string.h>
main()
{
    struct record {
        char name[2];
        char letter;
    };

    struct record student[10];
    strcpy(student[0].name,"t");//copy "t" to first struct's name variable
    strcpy(student[1].name,"ri");//copy "ri" to second struct's name variable
    student[0].letter='a';//copy "a" to first struct's letter variable
    student[1].letter='b';//copy "b" to second struct's letter variable
    printf("%s %s %c %c", student[0].name, student[1].name, student[0].letter, student[1].letter);
}

我期望的输出是：t ri a b

但是我得到：t rib a b

我做错了什么？

Answer 1

strcpy(student1.name,"ri");//copy "ri" to second struct's name variable

这绝对是错误的。 "ri" 实际上也是三个字符，因为空终止符也是如此。因此，您将三个字节复制到具有 2 个字节的数组，从而调用 undefined behaviour。

Answer 2

在您的代码中，name 是一个包含两个 char 的数组。 OTOH，"ri" 的大小是三个字符，包括空终止符。

所以，通过说

  strcpy(student[1].name,"ri");

你超出了内存。这会调用 undefined behavior.

来自man page

[...] The strings may not overlap, and the destination string dest must be large enough to receive the copy. [...] If the destination string of a strcpy() is not large enough, then anything might happen.

一旦命中UB，程序行为就无法自圆其说了。

为了能够保持"ri"，你需要改变

char name[2];

到

char name[3];

Answer 3

#include<stdio.h>
#include<string.h>
main()
{
    struct record {
        char name[3];
        char letter;
    };

    struct record student[10];
    strcpy(student[0].name,"t");//copy "t" to first struct's name variable
    strcpy(student[1].name,"ri");//copy "ri" to second struct's name variable
    student[0].letter='a';//copy "a" to first struct's letter variable
    student[1].letter='b';//copy "b" to second struct's letter variable
    printf("%s %s %c %c",student[0].name,student[1].name,student[0].letter,student[1].letter);
}

结构数组复制如何在 C 中工作？

How does structure array copying work in C?

c

arrays

structure

strcpy