c和perl之间的printf差异

Question

In C:

printf ("%u", 17860374324089702869);

prints "3336554965" (this is the answer I want!)

In perl, the same line of code prints "17860374324089702869".

How can I get perl to produce "3336554965" from "17860374324089702869".

My gut says this is 64-bit vs 32-bit issue, but it does not appear that the perl statement is doing anything at all to the number.

Answer 1

In C, printf %u takes an unsigned int . Whether that is a 32-bit or 64-bit value or something different entirely depends on your C implementation. The standard only requires that int be at least 16 bits wide. (Since the value you're passing in your example doesn't fit in an unsigned int , I'm pretty sure your code has undefined behavior: You're not passing an unsigned int to %u as it expects you to.)

In Perl, %u converts its argument to an unsigned integer. The width of this integer depends on how Perl was compiled; it is normally either a 32-bit or 64-bit type. You can query the configuration details for your perl by running perl -V:uvsize (the size of Perl's unsigned integer type in bytes) and perl -V:uvtype (the corresponding C type).

If you want to explicitly truncate your value to 32 bits, the easiest way is probably to apply a bit mask:

use constant { UINT32_MAX => 0xffff_ffff };
print 17860374324089702869 & UINT32_MAX, "\n";

Then you don't really need printf either.

Answer 2

There's no such unsigned 32-bit integer in Perl. The 17860374324089702869 become 3336554965 in C is because unsigned 32-bit integer has a range 0 to 4294967295, when there's overflow occur, the integer would came up from the lower bound and vice versa

eg

unsigned int a = 4294967296 // will be 0 because unsigned integer upper limit is 4294967295

So you only have to module it with 4294967296 because there's 4294967296 numbers from 0 to 4294967295

#!/usr/bin/perl

printf "%09u", 17860374324089702869 % 4294967296

Answer 3

This is generally not about Perl's printf / sprintf , but about how Perl stores integers and passes integers to functions. As mentioned in other responses, there is no 32-bit integer type in Perl.

Basically, you need to strip the half of your 64-bit integer to make it 32-bit. To do this, just use binary & :

perl -e 'printf("%09u\n", 17860374324089702869 & 0xffffffff);'
3336554965

Here, the & operation leaves bits of the lower 4 bytes unchanged, but sets bits of the higher 4 bytes to zeros.

Answer 4

The C language has a plethora of numeric data types (and sizes). Perl has 3 type of data -- " scalars, arrays of scalars, and associative arrays of scalars ",

In Perl, scalar values are polymorphic, so the value 17860374324089702869 can be treated as a string or a number, depending on how it is used. When used in a printf statement, it treated as the number 17860374324089702869 and printed as such.

In C, an unsigned integer is treated as an integer number in a range specified by the word size on the architecture you are compiling the program for. You should have received warning when you compiled. When I compiled the above statement with -Wall, here is the warning I got:

warning: format '%u' expects argument of type 'unsigned int', but argument 2 has type '__int128' [-Wformat=]

See this post discussing the maximum integer value that can be count to in Perl.