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Make || and && operators in constant expressions yield results of type int.

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Previously the result type was based on the operand types (using the arithmetic conversions), which is incorrect. The following program illustrates the issue:

#include <stdio.h>
int main(void)
{
    /* should print "1 0 2 2" */
    printf("%i %i %lu %lu\n", 0L || 2, 0.0 && 2,
           sizeof(1L || 5), sizeof(1.0 && 2.5));
}
This commit is contained in:
Stephen Heumann 2016-12-20 22:40:24 -06:00
parent 40bed0a93e
commit 280f67e846
1 changed files with 16 additions and 8 deletions
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24
Expression.pas
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@ -1039,10 +1039,14 @@ var
dispose(op^.left);
op^.left := nil;
case op^.token.kind of
barbarop : {||}
op1 := ord((op1 <> 0) or (op2 <> 0));
andandop : {&&}
op1 := ord((op1 <> 0) and (op2 <> 0));
barbarop : begin {||}
op1 := ord((op1 <> 0) or (op2 <> 0));
ekind := intconst;
end;
andandop : begin {&&}
op1 := ord((op1 <> 0) and (op2 <> 0));
ekind := intconst;
end;
carotch : op1 := op1 ! op2; {^}
barch : op1 := op1 | op2; {|}
andch : op1 := op1 & op2; {&}
@ -1147,10 +1151,14 @@ var
dispose(op^.left);
op^.left := nil;
case op^.token.kind of
barbarop : {||}
rop1 := ord((rop1 <> 0.0) or (rop2 <> 0.0));
andandop : {&&}
rop1 := ord((rop1 <> 0.0) and (rop2 <> 0.0));
barbarop : begin {||}
op1 := ord((rop1 <> 0.0) or (rop2 <> 0.0));
ekind := intconst;
end;
andandop : begin {&&}
op1 := ord((rop1 <> 0.0) and (rop2 <> 0.0));
ekind := intconst;
end;
eqeqop : begin {==}
op1 := ord(rop1 = rop2);
ekind := intconst;