C语言 简单的 int 到 char[] 转换
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Simple int to char[] conversion
提问by Waypoint
I have a simple question
我有一个简单的问题
How to simply convert integer (getting values 0-8) to char, e.g. char[2] in C?
如何简单地将整数（获取值 0-8）转换为 char，例如 C 中的 char[2]？
Thanks
谢谢
回答by Yoko Zunna
main()
{
int i = 247593;
char str[10];
sprintf(str, "%d", i);
// Now str contains the integer as characters
}
Hope it will be helpful to you.
希望对你有帮助。
回答by Prashant
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
void main()
{
int a = 543210 ;
char arr[10] ="" ;
itoa(a,arr,10) ; // itoa() is a function of stdlib.h file that convert integer
// int to array itoa( integer, targated array, base u want to
//convert like decimal have 10
for( int i= 0 ; i < strlen(arr); i++) // strlen() function in string file thar return string length
printf("%c",arr[i]);
}
回答by Erik
Use this. Beware of i's larger than 9, as these will require a char array with more than 2 elements to avoid a buffer overrun.
用这个。当心 i 大于 9，因为这些将需要一个包含 2 个以上元素的字符数组以避免缓冲区溢出。
char c[2];
int i=1;
sprintf(c, "%d", i);
回答by xanatos
You can't truly do it in "standard" C, because the size of an int and of a char aren't fixed. Let's say you are using a compiler under Windows or Linux on an intel PC...
你不能在“标准”C 中真正做到这一点，因为 int 和 char 的大小不是固定的。假设您在英特尔 PC 上使用 Windows 或 Linux 下的编译器...
int i = 5; char a = ((char*)&i)[0]; char b = ((char*)&i)[1];Remember of endianness of your machine! And that int are "normally" 32 bits, so 4 chars!
int i = 5; char a = ((char*)&i)[0]; char b = ((char*)&i)[1];记住你的机器的字节序！而那个 int “通常”是 32 位，所以是 4 个字符！
But you probably meant "i want to stringify a number", so ignore this response :-)
但是您可能的意思是“我想对数字进行字符串化”，因此请忽略此回复:-)
回答by ChrisJ
If you want to convert an int which is in the range 0-9 to a char, you may usually write something like this:
如果要将 0-9 范围内的 int 转换为 char，通常可以这样写：
int x;
char c = '0' + x;
Now, if you want a character string, just add a terminating '\0' char:
现在，如果你想要一个字符串，只需添加一个终止的 '\0' 字符：
char s[] = {'0' + x, '##代码##'};
Note that:
注意：
- You must be sure that the int is in the 0-9 range, otherwise it will fail,
- It works only if character codes for digits are consecutive. This is true in the vast majority of systems, that are ASCII-based, but this is not guaranteed to be true in all cases.
- 一定要确定int在0-9范围内，否则会失败，
- 仅当数字的字符代码是连续的时才有效。在绝大多数基于 ASCII 的系统中都是如此，但不能保证在所有情况下都如此。
