itoa函数将integer转换为以null终止的字符串。它也可以转换负数。 itoa函数的标准定义如下:-
char* itoa(int num, char* buffer, int base)
第三个参数库指定转换基数。例如:-如果base为2,则它将整数转换为其二进制兼容字符串;如果base为16,则它将创建整数的十六进制转换字符串形式。
如果base为10并且value为负,则结果字符串前面带有减号(-)。在任何其他基数下,值始终被认为是无符号的。
参考:http://www.cplusplus.com/reference/cstdlib/itoa/?kw=itoa
例子:
itoa(1567, str, 10) should return string "1567"
itoa(-1567, str, 10) should return string "-1567"
itoa(1567, str, 2) should return string "11000011111"
itoa(1567, str, 16) should return string "61f"
给定数字的各个数字必须经过处理,并且其相应的字符必须放在给定的字符串。使用给定基数的重复除法,我们得到从最低有效位到最高有效位的各个数字。但是在输出中,这些数字需要相反的顺序。因此,我们将经过重复除法后获得的字符串反转并返回。
/* A C++ program to implement itoa() */
#include
using namespace std;
/* A utility function to reverse a string */
void reverse(char str[], int length)
{
int start = 0;
int end = length -1;
while (start < end)
{
swap(*(str+start), *(str+end));
start++;
end--;
}
}
// Implementation of itoa()
char* itoa(int num, char* str, int base)
{
int i = 0;
bool isNegative = false;
/* Handle 0 explicitely, otherwise empty string is printed for 0 */
if (num == 0)
{
str[i++] = '0';
str[i] = '\0';
return str;
}
// In standard itoa(), negative numbers are handled only with
// base 10. Otherwise numbers are considered unsigned.
if (num < 0 && base == 10)
{
isNegative = true;
num = -num;
}
// Process individual digits
while (num != 0)
{
int rem = num % base;
str[i++] = (rem > 9)? (rem-10) + 'a' : rem + '0';
num = num/base;
}
// If number is negative, append '-'
if (isNegative)
str[i++] = '-';
str[i] = '\0'; // Append string terminator
// Reverse the string
reverse(str, i);
return str;
}
// Driver program to test implementation of itoa()
int main()
{
char str[100];
cout << "Base:10 " << itoa(1567, str, 10) << endl;
cout << "Base:10 " << itoa(-1567, str, 10) << endl;
cout << "Base:2 " << itoa(1567, str, 2) << endl;
cout << "Base:8 " << itoa(1567, str, 8) << endl;
cout << "Base:16 " << itoa(1567, str, 16) << endl;
return 0;
}
输出:
Base:10 1567
Base:10 -1567
Base:2 11000011111
Base:8 3037
Base:16 61f
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