atoi
Let’s take a look at the official documents ( portal )
interface
int atoi (const char * str);
Convert string to integer
- Parses the C-string str interpreting its content as an integral
number, which is returned as a value of type int. - The function first discards as many whitespace characters (as in
isspace) as necessary until the first non-whitespace character is
found. Then, starting from this character, takes an optional initial
plus or minus sign followed by as many base-10 digits as possible,
and interprets them as a numerical value. - The string can contain additional characters after those that form
the integral number, which are ignored and have no effect on the
behavior of this function. - If the first sequence of non-whitespace characters in str is not a
valid integral number, or if no such sequence exists because either
str is empty or it contains only whitespace characters, no conversion
is performed and zero is returned.
Parameters
- C-string beginning with the representation of an integral number.
Return Value
- On success, the function returns the converted integral number as an
int value. - If the converted value would be out of the range of representable
values by an int, it causes undefined behavior. See strtol for a more
robust cross-platform alternative when this is a possibility.
Speaking of torture, use code verification to verify
int main() {
//模拟实现atoi
char str[] = "-22675";
int ans = atoi(str);
printf("%d\n",ans);
system("pause");
return 0;
}
Operation result: -22675
int main() {
//模拟实现atoi
char str[] = " -22675";// 添加空格
int ans = atoi(str);
printf("%d\n",ans);
system("pause");
return 0;
}
Operation result: -22675
int main() {
//模拟实现atoi
char str[] = " +22675";
int ans = atoi(str);
printf("%d\n",ans);
system("pause");
return 0;
}
Running result: 22675
int main() {
//模拟实现atoi
char str[] = " ww +22675";
int ans = atoi(str);
printf("%d\n",ans);
system("pause");
return 0;
}
Operation result: 0
int main() {
//模拟实现atoi
char str[] = " +22675wer";
int ans = atoi(str);
printf("%d\n",ans);
system("pause");
return 0;
}
Running result: 22675
After trying a lot of torture, everyone 1 should have discovered his function. Realize atoi by yourself
C language implementation
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<assert.h>
#include<windows.h>
int my_atoi(const char * str) {
assert(str);
int j = 0;
while (str[j] != '\0') {
// 将字符串前边的空格跳过
while (str[j] != '\0' && str[j] == ' ') {
j++;
}
//如果出现+ - 或者数字就结束循环，否则返回0
if (str[j]!='+' && str[j] != '-' && (str[j]<'0' ||str[j]>'9')) {
return 0;
}
break;
}
// 程序走到这里说明开头是+123 或者-123类似开头
int ans = 0;
int tmp = 1;
// 对 + - 处理一下，利用tmp标记+ -
if (str[j] == '+') {
j++;
}
if (str[j] == '-') {
tmp = -1;
j++;
}
int count = 0;
// 处理数字
while (str[j] != '\0'&& str[j] >='0' &&str[j]<='9') {
//printf("%d", pow(10, count));
ans += pow(10,count)*(str[j] - '0');
count++;
j++;
}
int num = 0;
// 反转一下
while (ans) {
num += pow(10, --count)*(ans % 10);
ans = ans / 10;
}
return num*tmp;
}
int main() {
//模拟实现atoi
char str[] = " +22675wer";
int ans = my_atoi(str);
printf("%d\n",ans);
system("pause");
return 0;
}