How to change the eflags register value in GDB?

Question

set $eflags does not change eflags value.

The old eflags value remains after eg. =>$set $eflag=0x243 [this is just an example input].

Alternatively, is there any way to set individual flags of eflags?

I'm looking for something like: set ZF[zero flag]. Is there a gdb command to do that?

Answer 1

set $eflags without parenthesis works in GDB 7.7.1

To set an individual flag, use its index. E.g., ZF is the 6th bit, so we can set it with:

set $ZF = 6                 # define a GDB variable: no effect on registers
set $eflags |= (1 << $ZF)   # set bit 6 in EFLAGS, the ZF bit.

The same goes for all other bitwise operations: How do you set, clear, and toggle a single bit?

# Clear
set $eflags &= ~(1 << $ZF)

# Toggle
set $eflags ^= (1 << $ZF)

What causes confusion is that many bits are either reserved, cannot be modified directly by any instruction, or cannot be modified from user mode, see also: How to read and write x86 flags registers directly? and so GDB does not touch them.

For example:

(gdb) set $eflags = 0
(gdb) i r eflags
eflags         0x202    [ IF ]
(gdb) set $eflags = 0xFFFFFFFF
(gdb) i r eflags
eflags         0x54fd7  [ CF PF AF ZF SF TF IF DF OF NT RF AC ]

0x202 in binary is:

0010 0000 0010

0x54fd7 in binary is:

0101  0100 1111 1101 0111

TODO understand why each of those bits were set or not, by looking at the manual http://www.intel.com/content/dam/www/public/us/en/documents/manuals/64-ia-32-architectures-software-developer-vol-1-manual.pdf and GDB source code.

Ones that I understand:

• all reserved registers were left at their fixed value: 1 for bit 1, and 0 for bits 3, 5, 15 and 22-31