C++ const char* 应该在 C 中释放吗

发布于07月27日

请考虑以下代码.我在问这是否正确，是否遗漏了什么.还有别的办法吗.

#include <stdio.h>
#include <string.h>
const char *getOrder (const char *day)
{
    if (strncmp(day, "Monday", 7) == 0) {
        return "first";
    } else if (strncmp(day, "Tuesday", 7) == 0) {
        return "second";
    } else if (strncmp(day, "Wednesday", 9) == 0) {
        return "third";
    } else if (strncmp(day, "Thursday", 8) == 0) {
        return "forth";
    } else if (strncmp(day, "Friday", 6) == 0) {
        return "fifth";
    } else if (strncmp(day, "Saturday", 8) == 0) {
        return "sixth";
    } else if (strncmp(day, "Sunday", 6) == 0) {
        return "seventh";
    } else
        return NULL;
}
int main()
{
    const char* str = NULL;
    str = getOrder ("Monday");
    printf("str : %s\n", str);
    return 0;
}

推荐答案

if/else梯形图过于复杂，容易出现"复制/粘贴"错误.

似乎您需要一个"查找表"将工作日名称转换为不同的字符串.

const char *getOrder (const char *day) {
    char *LUT[][2] = {
    { "Monday", "first", },
    { "Tuesday", "second", },
    { "Wednesday", "third", },
    { "Thursday", "forth", },
    { "Friday", "fifth", },
    { "Saturday", "sixth", },
    { "Sunday", "seventh", },
    };
    
    for( i = 0; i < sizeof(LUT)/sizeof(LUT[0]); i++ )
        if( strncmp(day, LUT[i][0], strlen( LUT[i][0] ) ) == 0 )
            return LUT[ i ][ 1 ];

    return "NOT A DAY NAME";
}

它仍在反复搜索...

这里有一个很小的哈希函数，可以将字符串日名称快速转换为整数值(不区分大小写).玩它.也许你会try 将其直接改为看似合理的工作日名称.

// Convert a day name (or 3 letter abbrev.) to index (1-7, Sun=1). Beware false positives!
static int wkdayOrd( char cp[] ) { return "65013427"[*cp/2 + ~cp[1] & 0x7] & 0x7; }