integer-overflow

I want signed integers to overflow when they become too big. How do I achieve that without using the next biggest datatype (or when I am already at int128_t)? For example using 8bit integers 19*12 is commonly 260, but I want the result 1 11 10 01 00 with the 9th bit cut off, thus -27. Signed overflow is undefined in C, and that's for real . One solution follows: signed_result = (unsigned int)one_argument + (unsigned int)other_argument; The above solution involves implementation-defined behavior in the final conversion from unsigned to int but do not invoke undefined behavior. With most

I've made a simple program and compiled it with GCC 4.4/4.5 as follows: int main () { char u = 10; char x = 'x'; char i = u + x; return 0; } g++ -c -Wconversion a.cpp And I've got the following: a.cpp: In function ‘int main()’: a.cpp:5:16: warning: conversion to ‘char’ from ‘int’ may alter its value The same warning I've got for the following code: unsigned short u = 10; unsigned short x = 0; unsigned short i = u + x; a.cpp: In function ‘int main()’: a.cpp:5:16: warning: conversion to ‘short unsigned int’ from ‘int’ may alter its value Could anyone please explain me why addition of two chars

We have some integer arithmetic which for historical reasons has to work the same on PHP as it does in a few statically typed languages. Since we last upgraded PHP the behavior for overflowing integers has changed. Basically we are using following formula: function f($x1, $x2, $x3, $x4) { return (($x1 + $x2) ^ $x3) + $x4; } However, even with conversions: function f($x1, $x2, $x3, $x4) { return intval(intval(intval($x1 + $x2) ^ $x3) + $x4); } I am still ending up with the completely wrong number... For example, with $x1 = -1580033017, $x2 = -2072974554, $x3 = -1170476976) and $x4 = -1007518822

I looked into this stackoverflow question relating to Big Integer and specifically I do not understand this line (the words in italics): In the BigInteger class, I have no limits and there are some helpful functions there but it is pretty depressing to convert your beautiful code to work with the BigInteger class, specially when primitive operators don't work there and you must use functions from this class. I don't know what I am missing but to represent something that has no limit you would require infinite memory ? Whats is the trick here ? There is no theoretical limit. The BigInteger

I have read in many places that unsigned integer overflow is well-defined in C unlike the signed counterpart. Is underflow the same? For example: unsigned int x = -1; // Does x == UINT_MAX? Thanks. I can't recall where, but i read somewhere that arithmetic on unsigned integral types is modular, so if that were the case then -1 == UINT_MAX mod (UINT_MAX+1). Stephen Canon §6.2.5, paragraph 9: A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the