Aam*_*mir 69
atoi可以为你做到这一点
例:
char string[] = "1234";
int sum = atoi( string );
printf("Sum = %d\n", sum ); // Outputs: Sum = 1234
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使用atoi()来自 <stdlib.h>
http://linux.die.net/man/3/atoi
或者,编写atoi()将转换char*为的函数int
int a2i(const char *s)
{
int sign=1;
if(*s == '-'){
sign = -1;
s++;
}
int num=0;
while(*s){
num=((*s)-'0')+num*10;
s++;
}
return num*sign;
}
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