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perl-5.42.2
⭐ Starred 2013 / integer
Security Advisories (3)
CVE-2026-13221 (2026-07-13)

Perl versions through 5.43.9 produce silently incorrect regular expression matches when an alternation of more than 65535 fixed string branches is compiled into a trie in Perl_study_chunk. When such branches are combined into a trie, the delta between the first branch and the shared tail is stored in a 16-bit field. A branch count above 65535 overflows the field, and the trie's match decision table is truncated with no warning or error. A pattern of this shape produces false positive matches (matching strings it should not) and false negative matches (failing to match strings it should). When such a pattern gates an access or filtering decision, the result is wrong.

CVE-2026-8376 (2026-05-25)

Perl versions through 5.43.10 have a heap buffer overflow when compiling regular expressions with a repeated fixed string on 32-bit builds. Perl_study_chunk in regcomp_study.c checked the size of the joined substring buffer in characters rather than bytes. For a quantified fixed substring with a large minimum count, the byte length mincount * l could overflow SSize_t, producing an undersized SvGROW allocation; the subsequent copy writes past the end of the buffer. A caller that compiles an attacker-controlled regular expression on a 32-bit perl build triggers a heap buffer overflow at compile time.

CVE-2026-57432 (2026-07-13)

Perl versions through 5.43.10 have an integer overflow in S_measure_struct leading to an out-of-bounds heap read in pack and unpack. S_measure_struct adds each item's size times its repeat count to a running total with no overflow check, so a large repeat count in a pack or unpack template wraps the signed SSize_t total negative. The @, X, and x position codes then guard their moves with a signed length comparison that passes when the length is negative, advancing the buffer pointer out of bounds. A template derived from untrusted input can read heap memory past the buffer and return it to the caller.

NAME

integer - Perl pragma to use integer arithmetic instead of floating point

SYNOPSIS

use integer;
$x = 10/3;
# $x is now 3, not 3.33333333333333333

DESCRIPTION

This tells the compiler to use integer operations from here to the end of the enclosing BLOCK. On many machines, this doesn't matter a great deal for most computations, but on those without floating point hardware, it can make a big difference in performance.

Note that this only affects how most of the arithmetic and relational operators handle their operands and results, and not how all numbers everywhere are treated. Specifically, use integer; has the effect that before computing the results of the arithmetic operators (+, -, *, /, %, +=, -=, *=, /=, %=, and unary minus), the comparison operators (<, <=, >, >=, ==, !=, <=>), and the bitwise operators (|, &, ^, <<, >>, |=, &=, ^=, <<=, >>=), the operands have their fractional portions truncated (or floored), and the result will have its fractional portion truncated as well. In addition, the range of operands and results is restricted to that of familiar two's complement integers, i.e., -(2**31) .. (2**31-1) on 32-bit architectures, and -(2**63) .. (2**63-1) on 64-bit architectures. For example, this code

use integer;
$x = 5.8;
$y = 2.5;
$z = 2.7;
$a = 2**31 - 1;  # Largest positive integer on 32-bit machines
$, = ", ";
print $x, -$x, $x+$y, $x-$y, $x/$y, $x*$y, $y==$z, $a, $a+1;

will print: 5.8, -5, 7, 3, 2, 10, 1, 2147483647, -2147483648

Note that $x is still printed as having its true non-integer value of 5.8 since it wasn't operated on. And note too the wrap-around from the largest positive integer to the largest negative one. Also, arguments passed to functions and the values returned by them are not affected by use integer;. E.g.,

srand(1.5);
$, = ", ";
print sin(.5), cos(.5), atan2(1,2), sqrt(2), rand(10);

will give the same result with or without use integer; The power operator ** is also not affected, so that 2 ** .5 is always the square root of 2. Now, it so happens that the pre- and post- increment and decrement operators, ++ and --, are not affected by use integer; either. Some may rightly consider this to be a bug -- but at least it's a long-standing one.

Finally, use integer; also has an additional affect on the bitwise operators. Normally, the operands and results are treated as unsigned integers, but with use integer; the operands and results are signed. This means, among other things, that ~0 is -1, and -2 & -5 is -6.

Internally, native integer arithmetic (as provided by your C compiler) is used. This means that Perl's own semantics for arithmetic operations may not be preserved. One common source of trouble is the modulus of negative numbers, which Perl does one way, but your hardware may do another.

% perl -le 'print (4 % -3)'
-2
% perl -Minteger -le 'print (4 % -3)'
1

See "Pragmatic Modules" in perlmodlib, "Integer Arithmetic" in perlop