c - 带有隐式空字符的 sizeof() 和 strlen()

Question

如果我是对的，在数组c 中，\045 被视为单八进制，其对应的ASCII 字符是%。因此 sizeof(c) 是 4。

同样在数组d中，\099应该被当作单八进制值，对应的ASCII字符是R和 sizeof(d) 应该是 4，但不是。

您能否详细说明 sizeof(d) 和 sizeof(e) 的情况？

代码:

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

int main(void)
{
    char      a[] = {"12345"};
    char      b[] = {"12345\0"};

    char      c[] = {"12\045"};
    char      d[] = {"12\099"};
    char      e[] = {"12\0133"};

    printf("\ta         = %s  \n", a);
    printf("\tsizeof(a) = %ld \n", sizeof(a));
    printf("\tstrlen(a) = %ld \n", strlen(a));

    printf("\n");

    printf("\tb         = %s  \n", b);
    printf("\tsizeof(b) = %ld \n", sizeof(b));
    printf("\tstrlen(a) = %ld \n", strlen(b));

    printf("\n");
    printf("\tc         = %s  \n", c);
    printf("\tsizeof(c) = %ld \n", sizeof(c));
    printf("\tstrlen(c) = %ld \n", strlen(c));

    printf("\n");

    printf("\td         = %s  \n", d);
    printf("\tsizeof(d) = %ld \n", sizeof(d));
    printf("\tstrlen(d) = %ld \n", strlen(d));

    printf("\n");

    printf("\te         = %s  \n", e);
    printf("\tsizeof(e) = %ld \n", sizeof(e));
    printf("\tstrlen(e) = %ld \n", strlen(e));

    return (0);
}

实际结果:

    a         = 12345  
    sizeof(a) = 6 
    strlen(a) = 5 

    b         = 12345  
    sizeof(b) = 7 
    strlen(a) = 5 

    c         = 12%  
    sizeof(c) = 4 
    strlen(c) = 3 

    d         = 12  
    sizeof(d) = 6 
    strlen(d) = 2 

    e         = 12
                      3  
    sizeof(e) = 5 
    strlen(e) = 4

Answer 1

'9' 不是有效的八进制数字。字符串 "\099" 是一个 '\0' 后跟两个 '9' 字符。因此，字符串长度为 2，数组长度为 6。'\0' 是字符串终止符。

d[] 的初始化等价于:

char e[] = { '1','2','\0','9','9','\0' } ;