Convertion from U16 in U8 results overflow. Why?
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Hello all,
I made the following example:
typedef unsigned int U16;
typedef unsigned char U8;
// case 1;
U16 a = 0x0FFF;
U16 b = 0x0E00;
U8 c = 0x00;
//case 2:
U16 d = 0xFFFF;
U16 e = 0x0011;
U16 f = 0;
void main()
{ //case 1:
c = (U8) (a-b); // -> red check overflow
//case 2:
f = (U8) ((d + e)&0x00FF); // + red warning -> no effect of 0x00FF
}
When I run the Polyspace (Code Prover R2014b) analysis, I receive the following red checks:
1) Error: operation [conversion from unsigned int16 to unsigned int8] on scalar overflows (result is always strictly greater than MAX UINT8) conversion from unsigned int 16 to unsigned int 8
2) Error: operation [+] on scalar overflows (result is always strictly greater than MAX UINT16) operator + on type unsigned int 16
If I change the code from:
c = (U8) (a-b);
to
c = (U8) ((a-b)&0x00FF);
I don't receive the first red warning. Which is the correct configuration for overflow for no error occurring?
1 Commento
Titus Edelhofer
il 22 Apr 2015
Hi Christina,
is your question answered with the answers of Alex and myself?
Titus
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Più risposte (1)
Titus Edelhofer
il 10 Apr 2015
Modificato: Titus Edelhofer
il 10 Apr 2015
Hi,
hmm, I guess the first error is explainable: you indeed have an overflow here.
hex2dec('0FFF')-hex2dec('0E00')
ans =
511
which is larger than 255. This overflows when cast to U8.
The second error is also correct: your plus operation overflows in U16. Doing the "&" afterwards does not help here, because the error occurred inside the (d+e).
last but not least:
c = (U8) ((a-b)&0x00FF);
does not error, because (a-b) is fine (it's done in U16 and a>b). The "&" "removes" the bits larger than 255. Then casting to U8 is safe.
Titus
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