将 char 缓冲区转换为 c 结构

Question

在下面的代码中，我期待另一个输出! :

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct _cust {
        char    customerId[10];
        char    customerPhone[10];
        char    customerDep[4];
} cust;

int main(int argc, char **argv) {
        cust    *newCust;
        const char testChar[] = "11W35A5CT-012345678-CORP";

        newCust = (cust *)malloc(sizeof(struct _cust));
        newCust = (cust *)testChar;

        printf("CustomerId = %s\n", newCust->customerId);
        printf("CustomerPhone = %s\n", newCust->customerPhone);
        printf("CustomerDep = %s\n", newCust->customerDep);

        return 0;
}

输出是:

CustomerId = 11W35A5CT-012345678-CORP
CustomerPhone = 012345678-CORP
CustomerDep = CORP

我期待这个输出:

CustomerId = 11W35A5CT-
CustomerPhone = 012345678-
CustomerDep = CORP

谁能给我解释一下这是为什么？谢谢。

编辑:

为了避免混淆我的帖子，我在调试此程序时在此处添加了 gdb 跟踪:

(gdb) b main
Breakpoint 1 at 0x8048474: file name.c, line 11.
(gdb) run
Starting program: /home/evariste/src/customer_files/a.out 

Breakpoint 1, main (argc=1, argv=0xbffff2c4) at name.c:11
11  int main(int argc, char **argv) {
(gdb) n
13      const char testChar[] = "11W35A5CT-012345678-CORP";
(gdb) n
15      newCust = (cust *)malloc(sizeof(struct _cust));
(gdb) n
16      newCust = (cust *)testChar;
(gdb) n
21      printf("CustomerId = %s\n", newCust->customerId);
(gdb) print *newCust
$1 = {customerId = "11W35A5CT-", customerPhone = "012345678-", 
  customerDep = "CORP"}

所以我在这里看到 customerId = "11W35A5CT-"并且当我尝试 printf 时我得到了整个字符串？

Answer 1

printf() 将一直输出，直到遇到 \0 表示字符串结束为止。在任何连字符之后都没有 \0，因此 printf() 将从您给它的开始位置打印到 testChar.

此外，您泄漏了调用为您分配的 malloc 的内存。也许您想将字符串复制到结构中？