PHP:如何将bigint从int转换为string?

问题描述：

我希望能够将大整数转换为它们的全字符串导数.

I want to be able to convert big ints into their full string derivatives.

例如.

$bigint = 9999999999999999999;
$bigint_string = (string) $bigint;
var_dump($bigint_string);

输出

string(7) "1.0e+19"

但我需要

string(19) "9999999999999999999"

请不要告诉我，我最初应该将$ bigint值设置为字符串.那不是一个选择.我真的被困住了，不知道是否有可能?

Please don't tell me that I should just originally set the $bigint value as a string. That is not an option. I really am stuck and don't know if it is even possible?

答

您实际上应该确保输入您的意思:

You actually should ensure that you type what you mean:

$bigint = 9999999999999999999;

不是PHP整数，而是float:

Is not a PHP integer but float:

float(1.0E+19)

如果可以的话

$bigint = (int) 9999999999999999999;

您实际上已经设置了一个整数，但它不是您可能期望的数字:

You would have set an integer in fact, but it would not be the number you might have expected:

int(-8446744073709551616)

将其转换为字符串完全没问题，就像您可能已经猜到的那样.因此，在将数字写入代码时要小心，实际上要写出你的意思.

It's no problem at all to turn that into string as you might have guessed. So take care when you write numbers into code that you actually write what you mean.

再次查看此行:

$bigint = 9999999999999999999;

尝试了解您实际写的内容.它根本不是整数，因为PHP会将其转换为浮点数.请参见整数手册页中的示例#4在64位系统上的整数溢出 a>.

try to understand what you have actually written. It's not an integer at all because PHP will turn it into a float. See Example #4 Integer overflow on a 64-bit system in the integer manual page.

如果您需要更高的精度，请检出 GNU多精度，您正在寻找什么.但是，这不会改变在PHP中写数字的方式:

If you need higher precision, checkout GNU Multiple Precision, it might have what you're looking for. However, this won't change how to write numbers in PHP:

$bigint = gmp_init("9999999999999999999");
$bigint_string = gmp_strval($bigint);
var_dump($bigint, $bigint_string);

输出:

resource(4) of type (GMP integer)
string(19) "9999999999999999999"

PHP:如何将bigint从int转换为string?

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