integer-overflow

问题 I read a bit in C spec that unsigned variables(in particular unsigned short int ) perform some so called wrap around on integer overflow, although I couldn\'t find anything on signed variables except that I left with undefined behavior . My professor told me that their values also get wrapped around (maybe he just meant gcc). I thought the bits just get truncated and the bits I left with give me some weird value! What wrap around is and how is it different from just truncating bits. 回答1:

问题 What is an integer overflow error? Why do i care about such an error? What are some methods of avoiding or preventing it? 回答1: Integer overflow occurs when you try to express a number that is larger than the largest number the integer type can handle. If you try to express the number 300 in one byte, you have an integer overflow (maximum is 255). 100,000 in two bytes is also an integer overflow (65,535 is the maximum). You need to care about it because mathematical operations won't behave as

问题 System.out.println(Integer.MAX_VALUE + 1 == Integer.MIN_VALUE); is true. I understand that integer in Java is 32 bit and can\'t go above 2 31 -1, but I can\'t understand why adding 1 to its MAX_VALUE results in MIN_VALUE and not in some kind of exception. Not mentioning something like transparent conversion to a bigger type, like Ruby does. Is this behavior specified somewhere? Can I rely on it? 回答1: Because the integer overflows. When it overflows, the next value is Integer.MIN_VALUE .

问题 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

问题 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

问题 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

问题 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). 回答1: §6.2.5, paragraph 9: A computation involving unsigned operands can never overflow, because a result that cannot be represented by

问题 At first glance, this question may seem like a duplicate of How to detect integer overflow?, however it is actually significantly different. I\'ve found that while detecting an unsigned integer overflow is pretty trivial, detecting a signed overflow in C/C++ is actually more difficult than most people think. The most obvious, yet naive, way to do it would be something like: int add(int lhs, int rhs) { int sum = lhs + rhs; if ((lhs >= 0 && sum < rhs) || (lhs < 0 && sum > rhs)) { /* an overflow

问题 As we know, signed integer overflow is undefined behavior. But there is something interesting in C++11 cstdint documentation: signed integer type with width of exactly 8, 16, 32 and 64 bits respectively with no padding bits and using 2\'s complement for negative values (provided only if the implementation directly supports the type) See link And here is my question: since the standard says explicitly that for int8_t , int16_t , int32_t and int64_t negative numbers are 2\'s complement, is

问题 Unsigned integer overflow is well defined by both the C and C++ standards. For example, the C99 standard ( §6.2.5/9 ) states 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 largest value that can be represented by the resulting type. However, both standards state that signed integer overflow is undefined behavior. Again, from the C99