不要被别人的 memset(buf, -1, sizeof(buf)) 迷惑了 #2

district10 · 2016-08-11T12:46:42Z

有文件 test.c

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

int main()
{
    int m[2];
    printf( "sizeof(m): %lu\n", sizeof(m) );

    memset( m, -1, sizeof(m) );
    printf( "after memset(m,   -1, sizeof(m)): %20d (0x%08x), %20d (0x%08x)\n", m[0], m[0], m[1], m[1] );

    memset( m,  1, sizeof(m) );
    printf( "after memset(m,    1, sizeof(m)): %20d (0x%08x), %20d (0x%08x)\n", m[0], m[0], m[1], m[1] );

    memset( m, 0xFF, sizeof(m) );
    printf( "after memset(m, 0xFF, sizeof(m)): %20d (0x%08x), %20d (0x%08x)\n", m[0], m[0], m[1], m[1] );

    memset( m, 0xA5, sizeof(m) );
    printf( "after memset(m, 0xA5, sizeof(m)): %20d (0x%08x), %20d (0x%08x)\n", m[0], m[0], m[1], m[1] );
}

编译运行

$ gcc test.c -o test
$ ./test
sizeof(m): 8
after memset(m,   -1, sizeof(m)):                   -1 (0xffffffff),                   -1 (0xffffffff)
after memset(m,    1, sizeof(m)):             16843009 (0x01010101),             16843009 (0x01010101)
after memset(m, 0xFF, sizeof(m)):                   -1 (0xffffffff),                   -1 (0xffffffff)
after memset(m, 0xA5, sizeof(m)):          -1515870811 (0xa5a5a5a5),          -1515870811 (0xa5a5a5a5)

所以，要小心了。别人用 memset(buf, -1, sizeof(buf)) 来初始化 buf，你用 -2 可不是 -2 啊！

这涉及到 two's compliment 编码。

另一个不符合直觉的是 int buf[50] = { 0 }; 确实会把所有元素初始化为 0，但是 int buf[50] = { 3 }; 只回初始化第一个元素为 3，其它都是 0……至少在我的 gcc 和 g++ 上都是如此。

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不要被别人的 memset(buf, -1, sizeof(buf)) 迷惑了 #2

不要被别人的 memset(buf, -1, sizeof(buf)) 迷惑了 #2

district10 commented Aug 11, 2016 •

edited

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不要被别人的 memset(buf, -1, sizeof(buf)) 迷惑了 #2

不要被别人的 memset(buf, -1, sizeof(buf)) 迷惑了 #2

Comments

district10 commented Aug 11, 2016 • edited Loading

district10 commented Aug 11, 2016 •

edited

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