Thanks for the reply. I thought TREE_INT_CST_HIGH represented the high part of the integer when taken on 4 bits i.e 5 = 0|1|0|1 with high part = 0|1 and low part = 0|1. The high and low parts are then spread into 4 bits... Does your expectation mean that TREE_INT_CST_HIGHs purpose is solely as a sign indicator or does this indicate that 16 can also be repesented even though sixteen would not fit in HOST_BITS_PER_WIDE_INT bits? Regards, Primrose <From: Ian Lance Taylor [mailto:ian@xxxxxxxx] > >Why do you think TREE_INT_CST_HIGH would be 0|0|0|1 in your example? >I would expect >TREE_INT_CST_HIGH = 0|0|0|0 >TREE_INT_CST_LOW = 0|1|0|1 >Ian <Primrose.Mbanefo@xxxxxxxxxxxx> writes: >> I do not understand the gcc internal documentation on the integer_cst >> expression trees. >> An expression is given to calculate the value of the constant: >> >> ( (TREE_INT_CST_HIGH(e) << HOST_BITS_PER_WIDE_INT) >> + TREE_INT_CST_LOW(e)) >> >> Using the expression, what is wrong in the following example: >> >> Example >> ----------- >> Given integer = 5 >> HOST_BITS_PER_WIDE_INT = 4 //just an example HOST_WIDE_INT is a type >> for which the integers are HOST_BITS_PER_WIDE_INT bits long. >> >> Binary representation of given integer is 0|1|0|1 TREE_INT_CST_HIGH = >> 0|0|0|1 TREE_INT_CST_LOW = 0|0|0|1 >> ----------- >> >> This definitely does not yield 5 if I apply that expression, so, I >> imagine there is something wrong somewhere. >> If it does, then maybe I did not understand the expression? >> >> Could anybody please tell me where my thinking went wrong?
