gcc torture test pr52286.c

[Paul S]

I've ported gcc to a 16 bit CPU and have all torture tests 
passing bar one, pr52286.c

The offending lines of code are
  long a, b = 0;
  asm ("" : "=r" (a) : "0" (0));


which should cause zero to be assigned to the "a" SI sized variable.

Inspecting the generated code revealed that zero was only being 
assigned to the lower 16 bit half of "a".

ld    r2,0

I changed the inline asm statement to

  asm ("" : "=r" (a) : "0" (0L));

(note the change 0 to 0L) which caused the correct code to be 
generated ...

ld    r2,0
ld    r3,0

Curious, I performed an RTL dump and found that without the 
trailing 'L' following the '0'  the output of the expand pass 
looks like this ...

(insn 6 5 7 2 (set (reg:SI 29 [ a ])
        (asm_operands:SI ("") ("=r") 0 [
                (reg:HI 30)
            ]
             [
                (asm_input:HI ("0") 
../gcc/testsuite/gcc.c-torture/execute/pr52286.c:14)
            ]

compared to

(insn 6 5 7 2 (set (reg:SI 29 [ a ])
        (asm_operands:SI ("") ("=r") 0 [
                (reg:SI 30)
            ]
             [
                (asm_input:SI ("0") 
../gcc/testsuite/gcc.c-torture/execute/pr52286.c:14)
            ]

when 0L is used instead of just 0.

so it seems that the "0" constraint on the input operand is 
affecting the inferred mode of the output register operand ?

Am I reading this correctly or have I missed something ?

Thanks, Paul