C语言 如何在C中将整数转换为字符？
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How to convert integer to char in C?
提问by anik
How to convert integer to char in C?
如何在C中将整数转换为字符？
回答by Ofir
A char in C is already a number (the character's ASCII code), no conversion required.
C 中的字符已经是一个数字（字符的 ASCII 代码），不需要转换。
If you want to convert a digit to the corresponding character, you can simply add '0':
如果要将数字转换为相应的字符，只需添加“0”即可：
c = i +'0';
The '0' is a character in the ASCll table.
'0' 是 ASCll 表中的一个字符。
回答by ratty
You can try atoi() library function. Also sscanf() and sprintf() would help.
您可以尝试 atoi() 库函数。sscanf() 和 sprintf() 也会有所帮助。
Here is a small example to show converting integer to character string:
这是一个小示例，用于显示将整数转换为字符串：
main()
{
int i = 247593;
char str[10];
sprintf(str, "%d", i);
// Now str contains the integer as characters
}
Here for another Example
这是另一个例子
#include <stdio.h>
int main(void)
{
char text[] = "StringX";
int digit;
for (digit = 0; digit < 10; ++digit)
{
text[6] = digit + '0';
puts(text);
}
return 0;
}
/* my output
String0
String1
String2
String3
String4
String5
String6
String7
String8
String9
*/
回答by Amarghosh
Just assign the intto a charvariable.
只需将 分配int给一个char变量。
int i = 65;
char c = i;
printf("%c", c); //prints A
回答by Deepak Yadav
To convert integer to char only 0 to 9 will be converted. As we know 0's ASCII value is 48 so we have to add its value to the integer value to convert in into the desired character hence
要将整数转换为字符，只有 0 到 9 会被转换。我们知道 0 的 ASCII 值是 48，所以我们必须将它的值添加到整数值中以转换为所需的字符，因此
int i=5;
char c = i+'0';
回答by Anurag Semwal
To convert int to char use:
要将 int 转换为 char 使用：
int a=8;
char c=a+'0';
printf("%c",c); //prints 8
To Convert char to int use:
要将 char 转换为 int 使用：
char c='5';
int a=c-'0';
printf("%d",a); //prints 5
回答by kannadasan
void main ()
{
int temp,integer,count=0,i,cnd=0;
char ascii[10]={0};
printf("enter a number");
scanf("%d",&integer);
if(integer>>31)
{
/*CONVERTING 2's complement value to normal value*/
integer=~integer+1;
for(temp=integer;temp!=0;temp/=10,count++);
ascii[0]=0x2D;
count++;
cnd=1;
}
else
for(temp=integer;temp!=0;temp/=10,count++);
for(i=count-1,temp=integer;i>=cnd;i--)
{
ascii[i]=(temp%10)+0x30;
temp/=10;
}
printf("\n count =%d ascii=%s ",count,ascii);
}
