[Bug target/31006] long double constant is read as double in i386

2009-09-17 Thread ubizjak at gmail dot com
--- Comment #6 from ubizjak at gmail dot com 2009-09-17 12:46 --- Your d_a is defined as double, so gcc truncates constant to fit in the "double" value range. After that, you pass this double value as an argument to %Lf, which is wrong. You should write: printf("long double - do

[Bug target/31006] long double constant is read as double in i386

2007-03-01 Thread rguenth at gcc dot gnu dot org
--- Comment #5 from rguenth at gcc dot gnu dot org 2007-03-01 12:27 --- With gcc 4.1.2 and your testcase fixed to use %f for the double argument printf I get [EMAIL PROTECTED]:/tmp> gcc -o t t.c -m32 [EMAIL PROTECTED]:/tmp> ./t long double = 3.141592653589793238512808959

[Bug target/31006] long double constant is read as double in i386

2007-03-01 Thread rguenth at gcc dot gnu dot org
--- Comment #4 from rguenth at gcc dot gnu dot org 2007-03-01 12:22 --- What is the expected output of your testcase? -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=31006