C ++基本转换 [英] C++ base conversion

问题描述

您好，我正在尝试将一些代码从Windows移植到Linux.我有这个:

Hello I'm trying to port some code from Windows to Linux. I have this:

itoa(word,aux,2);

但是GCC无法识别itoa.我该如何以C ++方式转换为以2为基础?

But GCC doesn't recognize itoa. How can I do this conversion to base 2 on a C++ way?

谢谢;)

推荐答案

此处一些帮助

/* itoa:  convert n to characters in s */
 void itoa(int n, char s[])
 {
     int i, sign;

     if ((sign = n) < 0)  /* record sign */
         n = -n;          /* make n positive */
     i = 0;
     do {       /* generate digits in reverse order */
         s[i++] = n % 10 + '0';   /* get next digit */
     } while ((n /= 10) > 0);     /* delete it */
     if (sign < 0)
         s[i++] = '-';
     s[i] = '\0';
     reverse(s);
 }

您应根据需要对其进行调整(请注意，此参数有2个参数，而不是3个) 还要注意，反向功能也位于Wikipedia链接中.

You should adapt it to your needs (notice this one has 2 arguments, not 3) Also notice the reverse function is also in the wikipedia link.

此外，此处还有其他一些情况(但不适用于基本情况) 2)

Additionally, here are some other cases (but not for base 2)

此功能未在ANSI-C中定义 并且不是C ++的一部分，但是 一些编译器支持.

This function is not defined in ANSI-C and is not part of C++, but is supported by some compilers.

符合标准的替代方案 某些情况下可能是sprintf:

A standard-compliant alternative for some cases may be sprintf:

sprintf(str,"%d",value)转换为十进制基数.

sprintf(str,"%d",value) converts to decimal base.

sprintf(str,"%x",value)转换为十六进制基数.

sprintf(str,"%x",value) converts to hexadecimal base.

sprintf(str,"%o",value)转换为八进制.

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