C - 模 (%)，如何计算 IntToString 转换的值，其中 num/= 10

Question

public static final int MAX_DIGITS = 10;

String intToStr( int num ){
    int i = 0;
    boolean isNeg = false;
    char[] temp = new char[ MAX_DIGITS + 1 ];

    if( num < 0 ){
        num *= -1;
        isNeg = true;
    }

    // Fill buffer with digit characters in reverse order 
    while(num != 0){
        temp[i++] = (char)((num % 10) + ‘0’);
        num /= 10;
    }

    StringBuffer b = new StringBuffer();
    if( isNeg )
        b.append( ‘-’ );

    while( i > 0 ){
        b.append( temp[--i] );
    }

    return b.toString();
}

我能否获得一些帮助来理解以下代码行:

    // Fill buffer with digit characters in reverse order 
    while(num != 0){
        temp[i++] = (char)((num % 10) + ‘0’);
        num /= 10;
    }

据我了解，如果 num = 732，且 i = 0:

temp [0] = "2"; //732 % 10 = 73.2 so 2 is the remainder?
num = 73.2; //732 divided by 10
temp [1] = "32"; //73.2 % 10 = 7.32 so 32 is the remainder?
num = 7.32; //73.2 divided by 10
temp [2] = "732"; //7.32 % 10 = 0.732 so 732 is the remainder?
num = 0.732; //7.32 divided by 10

num 现在小于零，所以继续。

然而，现在我们正在倒退:

    while( i > 0 ){
        b.append( temp[--i] );
    }

但是我从物理计算中看到的是:

String str = temp[2] + temp[1] + temp[0]; 
str = "732" + "32" + "2";

实际发生的是 temp 正在存储:

    temp[2] = 7; 
    temp[1] = 3;
    temp[0] = 2; 

获取:

   str = "732";

但我不明白 temp 如何只得到一个数字。

Answer 1

那么让我们运行一下您的程序。

// Fill buffer with digit characters in reverse order 
while(num != 0){
    temp[i++] = (char)((num % 10) + ‘0’);
    num /= 10;
}

When i is 0 and num is 732   
first case
temp[0] = (char)((732%10) +'0');
temp[0] now has the value of 2
next line 732/=10
num = 73
tem[1] = (char)((73%10 +'0');
temp [1] now has the value of 3  
next line 73/=10
temp[2] = (char)((7%10 +'0');
temp[2] now has the value of 7