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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.
The integer type is used to define objects whose value is always a whole number. VHDL doesn't specify the exact number of bits for the integer type, but any VHDL implementation should support at least a 32-bit realization. We can specify the range of values that an object of type integer is going to have.
A VHDL integer is defined from range -2147483648 to +2147483647.
What is an integer overflow? An integer overflow or wraparound happens when an attempt is made to store a value that is too large for an integer type. The range of values that can be stored in an integer type is better represented as a circular number line that wraps around.
I was wondering if integer overflow is defined in VHDL. I wasn't able to find anything in the 2002 Specification.
As an example (Note, this might not compile, it's just a generic example...):
entity foo is port (
clk : std_logic
);
end entity;
architecture rtl of foo is
signal x : integer range 0 to 2 := 0;
begin
process (clk)
begin
if rising_edge(clk) then
x <= x + 1;
end if;
end process;
end architecture;
It's clear that x will go from 0 to 1, and then to 2. Is it defined what will happen on the next increment? Is that undefined behavior?
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