C语言中0、‘\0‘、‘0‘、NULL以及类型转化

#include<stdio.h>

int main()
{
	int a = '\0';//将'\0'的ASCII码值赋值给a
	char b = 0;//将0的二进制赋值给b
	int c = '0';//将字符'0'的ASCII码值赋值给c
	int d = a + b + c;//在相加时，由于b是char型，所以会整形提升为int型
	printf("%d\n", d);//以%d形式打印d内存空间的内容

	printf("%c\n", '0');//打印字符'0'
	printf("%d\n", '0');//打印字符'0'的ASCII码值
	printf("%c\n", 0); //空（即NULL）
	printf("%d\n", 0);//数字0
	printf("%c\n", 0 + '\0');//空（NULL） '\0'先转换成int,然后ASCII编码相加为0，
	//%c输出，转换成char,所以输出NULL
	printf("%d\n", 0 + '\0');//先转换成int，然后ASCII编码相加为0,%d输出，不用转换，输出数字0
	printf("%c\n", 0 + '0');//'0'先转换成int,然后ASCII编码相加为0，
	//%c输出，ASCII转换成char,所以输出字符'0'
	printf("%d\n", 0 + '0');//输出字符'0'的ASCII码值
}

#include<stdio.h>
int my_atoi(char* str)
{
	int i = 0;
	int flag = 1;//默认字符串为正
	int value = 0;
	if ('-' == str[0])
	{
		flag = -1;//字符串第一个字符为‘-',则将flag置为-1
		i++;
	}
	else if ('+' == str[0])//字符串第一个字符为'+'，则不改变flag
	{
		i++;
	}
	else//字符串不加符号，默认为正
	{
		;
	}
	while (str[i] >= '0' && str[i] <= '9')
	{
		value = value * 10 + str[i] - '0';
		i++;
	}

	return flag * value;
}

int main(void)
{
	char str1[] = "+12345";	
	char str2[] = "-12345";
	char str3[] = "12345";
	int a = my_atoi(str1);
	int b = my_atoi(str2);
	int c = my_atoi(str3);
	printf("%d\n", a);
	printf("%d\n", b);
	printf("%d\n", c);
	return 0;
}

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
void split(char* str, double* position)
{
        char* words[3];
        char* temp;
        temp = strtok(str, " ");
        int i = 0;
        while(temp!=NULL)
            {
                words[i] = temp;
                temp = strtok(NULL, " ");
                position[i] = atof(words[i]);
                i++;
            }
}
int main()  
{  
 
    char lineBuf[] = "11.12 11.13 11.14";
    std::cout << "1" << std::endl;  
    double position[3];
    split(lineBuf, position);
    std::cout << "1" << std::endl;
    for(int i=0; i<3; i++)
    {
//        std::cout << words[i] << std::endl;
//        position[i] = atof(words[i]);
        std::cout << position[i] << std::endl;        
    }
    return 0;  
}