string.c

// string.c
/*
纯粹的用户态的计算,实现较为简单
*/
#include "minicrt.h"
char* itoa_(long n, char* str, long radix) {
    char digit[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    char* p      = str;
    char* head   = str;
    if (!p || radix < 2 || radix > 36)  // radix代表是几进制
        return p;

    if (radix != 10 && n < 0)
        return p;

    if (n == 0) {  // 如果要转换的数字n为0，则直接在输出字符串中直接输出
        *p++ = '0';
        *p   = 0;
        return p;
    }

    if (radix == 10 && n < 0) {  // 如果是10进制，且为负数，则先添加负号，然后转正留待后续处理
        *p++ = '-';
        n    = -n;
    }

    while (n) {
        *p++ = digit[n % radix];
        n /= radix;
    }

    *p = 0;  // 数字转换完了，末尾添加0

    for (--p; head < p; ++head, --p) {  // 上面的数字字符串是倒序的，这里将数字字符串倒过来
        char temp = *head;
        *head     = *p;
        *p        = temp;
    }

    return str;
}

char* utoa_(unsigned long n, char* str, long radix) {
    char digit[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    char* p      = str;
    char* head   = str;

    if (!p || radix < 2 || radix > 36)
        return p;

    if (n == 0) {
        *p++ = '0';
        *p   = 0;
        return p;
    }

    while (n) {
        *p++ = digit[n % radix];
        n /= radix;
    }

    *p = 0;

    for (--p; head < p; ++head, --p) {
        char temp = *head;
        *head     = *p;
        *p        = temp;
    }

    return str;
}

long strcmp(const char* src, const char* dst) {
    long ret          = 0;
    unsigned char* p1 = (unsigned char*)src;
    unsigned char* p2 = (unsigned char*)dst;

    while (!(ret = *p1 - *p2) && *p2)
        ++p1, ++p2;
    if (ret < 0)
        ret = -1;
    else if (ret > 0)
        ret = 1;

    return (ret);
}

char* strcpy(char* dest, const char* src) {
    char* ret = dest;
    while (*src)
        *dest++ = *src++;
    *dest = '\0';  // 字符串拷贝完后，手动在末尾添加\0
    return ret;    // 返回copy后的字符串的首
}

unsigned long strlen_(const char* str) {
    long cnt = 0;
    if (!str)
        return 0;
    for (; *str != '\0'; ++str)
        ++cnt;  // printf("inside the for-loop %d\n", cnt);

    return cnt;
}