我正在尝试打印我的char数组，但它只是打印数字为什么？ [英] I am trying to print my char array but it just prints number why?

问题描述

#include <stdio.h>
#include <stdlib.h>

/* run this program using the console pauser or add your own getch, system("pause") or input loop */

int main(int argc, char *argv[])
{
int counter;
char car[3]={'a','b','c',};
counter = 0; 
 while(counter < 3)  
{

   printf("%d\n",car[counter]); 
   counter++; 
    } 
 
 

	
	return 0;

What I have tried:

i have tried changing the values and trying different codes but nothing comes up it just gives me ramdon numbers on the output which i find annoying cuz the values are ABC

推荐答案

从
改变

Change from

Quote：

printf（％d \ n，car [counter]）;

printf("%d\n",car[counter]);

to

printf("%c\n",car[counter]);

查看 printf文档 [ ^ ]。



顺便说一下，你得到的数字不是随机的。它们是 ASCII代码 [ ^ ]对应字符'a'，'b'，'c'。

Have a look at printf documentation[^].

By the way, the numbers you get are not random ones. They are the ASCII codes[^] corresponding to the characters 'a', 'b', 'c'.

因为您使用的是％d格式说明符 - 其中说打印我整数。它打印的数字是每个字符的ASCII表示。

试试这个：

Because you are using the "%d" format specifier - which says "print me an integer number". The number it is printing is the ASCII representation of each character.
Try this:

printf("%c\n",car[counter]);

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