Because 2147483648 is a long value as it does not fit in an int (in common system with 32-bit int and 64-bit long, on system with 32-bit long it is of type long long). So -2147483648 is of type long, not int. Remember in C, an unsuffixed decimal integer constant is of the first type … Read more
Because the int value is promoted to an unsigned int. specifically 0xFFFFFFFC on a 32-bit machine, which as an unsigned int is 4294967292, considerably larger than 10 C99 6.3.1.1-p2 If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; … Read more
This is quite subtle. Every integer literal in your program has a type. Which type it has is regulated by a table in 6.4.4.1: Suffix Decimal Constant Octal or Hexadecimal Constant none int int long int unsigned int long long int long int unsigned long int long long int unsigned long long int If a … Read more
For iterating backwards see this answer. Iterating forwards is almost identical. Just change the iterators / swap decrement by increment. You should prefer iterators. Some people tell you to use std::size_t as the index variable type. However, that is not portable. Always use the size_type typedef of the container (While you could get away with … Read more
Unsigned can hold a larger positive value and no negative value. Yes. Unsigned uses the leading bit as a part of the value, while the signed version uses the left-most-bit to identify if the number is positive or negative. There are different ways of representing signed integers. The easiest to visualise is to use the … Read more
I had the same problem and I figured there wasn’t any helpful answer around on the internet so hoping this should help everyone in the future. When using GoogleMaps for Android, you need two keys – debug and release. The “debug” key is kind of a misleading term. This key is also to be used … Read more
No, your 2nd code isn’t correct, but you are close: if you set int half = INT_MAX/2; int half1 = half + 1; the result of an addition is INT_MAX. (INT_MAX is always an odd number). So this is valid input. But in your routine you will have INT_MAX – half == half1 and you … Read more
For the official answer – Section 4.7 conv.integral “If the destination type is unsigned, the resulting value is the least unsigned integer congruent to the source integer (modulo 2n where n is the number of bits used to represent the unsigned type). [ Note: In a two’s complement representation, this conversion is conceptual and there … Read more
When comparing signed with unsigned, the compiler converts the signed value to unsigned. For equality, this doesn’t matter, -1 == (unsigned) -1. For other comparisons it matters, e.g. the following is true: -1 > 2U. EDIT: References: 5/9: (Expressions) Many binary operators that expect operands of arithmetic or enumeration type cause conversions and yield result … Read more
Binary operations between different integral types are performed within a “common” type defined by so called usual arithmetic conversions (see the language specification, 6.3.1.8). In your case the “common” type is unsigned int. This means that int operand (your b) will get converted to unsigned int before the comparison, as well as for the purpose … Read more