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Usually, INT_MIN is -2 ^ n and INT_MAX is 2 ^ n - 1
Is it guaranteed, that if x is positive number of type int then expressoin -x didn't cause overflow?
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Write a “C” function, int addOvf(int* result, int a, int b) If there is no overflow, the function places the resultant = sum a+b in “result” and returns 0. Otherwise it returns -1. The solution of casting to long and adding to find detecting the overflow is not allowed.
"Signed integer overflow" means that you tried to store a value that's outside the range of values that the type can represent, and the result of that operation is undefined (in this particular case, your program halts with an error).
Division. Division is between two operands of arithmetic type. Overflow can occur during two's complement signed integer division when the dividend is equal to the minimum (negative) value for the signed integer type and the divisor is equal to −1 . Division operations are also susceptible to divide-by-zero errors.
The - (unary minus) operator negates the value of the operand. The operand can have any arithmetic type. The result is not an lvalue. For example, if quality has the value 100 , -quality has the value -100 .
It is implicitly guaranteed, since it is true for all the allowed forms of signedness:
(examples with 16 bit int)
INT_MIN = -32767, INT_MAX = 32767
INT_MIN = -32768, INT_MAX = 32767
INT_MIN = -32767, INT_MAX = 32767
No other forms are allowed. As we can see, abs(INT_MIN) >= abs(INT_MAX) for all the allowed forms.
As a side note, INT_MAX is not allowed to be smaller than 32767 and INT_MIN is not allowed to be smaller than -32767. This is guaranteed by the requirements for limits.h.
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