| atoi | function |
int atoi ( const char * str ); | <cstdlib> |
Convert string to integer
Parses the C string str interpreting its content as an integral number, which is returned as an int value.
The function first discards as many whitespace characters 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 numerical 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.
Parameters
-
str
- 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 no valid conversion could be performed, a zero value is returned.
If the correct value is out of the range of representable values, INT_MAX or INT_MIN is returned.
Example
/* atoi example */
#include <stdio.h>
#include <stdlib.h>
int main ()
{
int i;
char szInput [256];
printf ("Enter a number: ");
fgets ( szInput, 256, stdin ); i = atoi (szInput); printf ("The value entered is %d. The double is %d./n",i,i*2);
return 0;
} |
Output:
Enter a number: 73
|
See also
| atol | Convert string to long integer (function) |
| atof | Convert string to double (function) |
| strtol | Convert string to long integer (function) |
转载自:http://www.cplusplus.com/reference/clibrary/cstdlib/atoi.html
本文详细介绍了C/C++中atoi函数的功能及使用方法。atoi用于将字符串转换为整数,首先会跳过字符串开头的所有空白字符,然后读取一个可选的正负号,随后读取尽可能多的数字字符并将其解释为一个数值。
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