c语言atoi与itoa用法以及超过long long int 型变量整数加减法

Question

那位能告诉一下这两个函数的具体使用方法和大位数整数加减法的做法

Answer 1

大整数的计算可以通过模运算来进行简化，也可以将其作为字符串读入，然后分段运算之后再进行拼接。

附：
atoi和itoa的示范代码如下：

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

void main( void )
{
char *s; int ix;
char buffer[20];
int i = 3445;
long l = -344115L;
unsigned long ul = 1234567890UL;

s = " -9885 pigs";
/* Test of atoi */
ix = atoi( s );
printf( "atoi test: ASCII string: %s\t\tinteger: %d\n", s, ix );

_itoa( i, buffer, 10 );
printf( "String of integer %d (radix 10): %s\n", i, buffer );
_itoa( i, buffer, 16 );
printf( "String of integer %d (radix 16): 0x%s\n", i, buffer );
_itoa( i, buffer, 2 );
printf( "String of integer %d (radix 2): %s\n", i, buffer );

_ltoa( l, buffer, 16 );
printf( "String of long int %ld (radix 16): 0x%s\n", l,

buffer );

_ultoa( ul, buffer, 16 );
printf( "String of unsigned long %lu (radix 16): 0x%s\n", ul,

buffer );
}

输出结果：

atoi test: ASCII string:
-9885 pigs

integer: -9885
String of integer 3445 (radix 10): 3445
String of integer 3445 (radix 16): 0xd75
String of integer 3445 (radix 2): 110101110101
String of long int -344115 (radix 16): 0xfffabfcd
String of unsigned long 1234567890 (radix 16): 0x499602d2

Answer 2

将字符串转换成整形数值，如果可以的话。
如：
char
*buf="1234";
int
i=atoi(buf);
//则i
=
1234;