C语言atoi()和itoa()函数的实现

makefile

.SUFFIXES: .c .o

CC=gcc
SRCS=test.c

OBJS=$(SRCS:.c=.o)
EXEC=test

all: $(OBJS)
	$(CC) -o $(EXEC) $(OBJS)
	@echo '-------------ok--------------'

.c.o:
	$(CC) -Wall -g -o $@ -c $< 

clean:
	rm -f $(OBJS)
	rm -f core*

myitoa

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

void resver(char *s)//反转字符串
{
	int len = strlen(s);
	//printf("len=%d\n",len);
	int i = 0;
	char tmp = 0;
	for (; i<len/2; i++)
	{
		tmp = s[i];
		s[i] = s[len-1-i];
		s[len-1-i] = tmp;
	}
}

const char *myitoa(int n)
{
	static char buf[100];//必须为static,或者是全局变量
	memset(buf, 0, sizeof(buf));
	int FlagF = 0;    //1-f, 0-z
	int i = 0;
	if (n < 0)
	{
		FlagF = 1;
		n = 0 - n;
	}
	while (n)
	{
		buf[i++] = n % 10 + '0';
		n = n / 10;
	}
	if (FlagF == 1)
	{
		buf[i++] = '-';
	}
	buf[i] = '\0';
	printf("before buf:%s\n", buf);
	resver(buf);
	printf("after buf:%s\n", buf);
	return buf;
}

int main()
{
	printf("%s\n", myitoa(123));
	printf("%s\n", myitoa(-123));
	return 0;
}

myatoi

#include <stdio.h>
int mystrlen(const char *s)
{
    int i = 0;
    while (*s++)
    {
        i++;
    }
    return i;
}
int power10(int n)
{
    if (n == 0)
    {
        return 1;
    }
    int val = 1;
    int i = 0;
    for (; i < n; i++)
    {
        val = val * 10;
    }
    return val;
}
int char2i(char c)
{
    if ((c >= '0') || (c <= '9'))
    {
        return (c - '0');
    } else
    {
        return 0;
    }
}
int myatoi(const char *str)
{
    int FlagF = 0; //1-f, 0-z
    const char *p = str;
    int i = 0;
    int value = 0;
    if (str[0] == '-')
    {
        FlagF = 1; //
        p = ++str;
    }
    int len = mystrlen(p);
    int lentmp = len;
    for (; i < len; i++)
    {
        value += char2i(p[i]) * power10(lentmp - 1);
        lentmp--;
    }
    if (FlagF == 1)
    {
        return (0 - value);
    } else
    {
        return value;
    }
}
int main()
{
    const char *s = "-123";
    printf("%d\n", myatoi(s));
    return 0;
}