如何将2d字符数组转换为2d int数组？ [英] How to convert a 2d char array to a 2d int array?

问题描述

  char puzzle [i] [j]; 
 int i，j，count = 0; 
 char value [81]; 
 
 for（i = 0; i <9; i ++）{
 for（j = 0; j< 9; j ++）{
 cin> value [count]; 
 puzzle [i] [j] = value [count]; 
 count ++; 
}} 
  

这是我到目前为止。我尝试使用atoi，但我需要一个char * str。
输入是：..4545 ..（数字和句点）

我试图将字符拼图[i] [j]转换为int拼图[i] [j] 。 char数组目前拥有..4545 ..，我想隐藏它，所以它只保存整数00454500。

解决方案

p> 参与精神调试器

执行Jedi的技巧...
$ b

这是你想要的代码：

  int puzzle [9] [9] // changed type 
 int i，j，count = 0; 
 char value [81]; 
 
 for（i = 0; i <9; i ++）{
 for（j = 0; j< 9; j ++）{
 cin> value [count]; 
 puzzle [i] [j] = value [count]  - '0'; //从ASCII数字转换为整数
 count ++; 
} 
} 
  

char puzzle[i][j];  
int i,j,count=0;  
char value[81];

for( i = 0; i < 9; i++){  
  for( j = 0; j < 9; j++){  
    cin >> value[count];  
    puzzle[i][j] = value[count];  
    count++;  
  }}

This what I have so far. I tried using atoi but I needed a char * str. The input is: ..4545.. (numbers and periods)
I'm trying to convert the char puzzle[i][j] to an int puzzle[i][j]. The char array currently holds "..4545.." and I want to covert it so it holds just integers "00454500".

解决方案

Engaging psychic debugger...

Performing Jedi mind tricks...

This is the code you want:

int puzzle[9][9];  // changed type
int i,j,count=0;  
char value[81];

for( i = 0; i < 9; i++ ) {  
  for( j = 0; j < 9; j++ ) {  
    cin >> value[count];  
    puzzle[i][j] = value[count] - '0';  // convert from ASCII digit to integer
    count++;  
  }
}

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