Not equal to or greater than c code#
Not equal to or greater than c free#

However this will incorrectly handle NaNs. This was relevant in the era of sub-megahertz processor speed and 1:1 CPU-to-memory speed ratios, but it is almost totally irrelevant today.

So, on some machines, using a 'less than' comparison might save one machine instruction. Implementation of 'if (A <= B) goto address 'īzs address also, Branch if the Zero bit is Set So, implementing a branch for A < B can be done in one instruction, because the carry bit is clear only in this case,, that is, Implementation of 'if (A < B) goto address 'īcz address Branch if Carry is Zero to the new addressīut, if we want to do a less-than-or-equal comparison, we need to do an additional check of the zero flag to catch the case of equality. The root issue is that integer comparison is inherently implemented via integer subtractions. Historically (we're talking the 1980s and early 1990s), there were some architectures in which this was true. Setae al Set al if above or equal (CF=0). Seta al Set al if above (CF=0 and ZF=0).įucomip st, st(1) (same thing as above)

Not equal to or greater than c code#

This holds true for x87 floating point as well: (Pretty much same code as above, but with double instead of int.) fld QWORD PTR įucomip st, st(1) Compare ST(0) and ST(1), and set CF, PF, ZF in EFLAGS There is then, no reason that an instruction testing two bits should take any more or less time than one testing only one (Ignoring gate propagation delay, which is much less than the clock period.) If one thinks about the actual circuitry used to implement the instructions, one can assume that there would be simple AND/OR gates on the different bits in EFLAGS, to determine whether the conditions are met. So, nothing in the Intel docs ever treats one Jcc instruction any differently from the others. When branches are predicted successfully, the latency of jcc is effectively zero. Selection of conditional jump instructions should be based on the recommendation of section Section 3.4.1, “Branch Prediction Optimization,” to improve the predictability of branches.The values for Jcc are: Latency Throughput For many instructions, the throughput of an instruction can be

Not equal to or greater than c free#

Wait before the issue ports are free to accept the same instructionĪgain. Throughput - The number of clock cycles required to Latency - The number of clock cycles that are required for theĮxecution core to complete the execution of all of the μops that form The same grouping is made together under the Optimization Reference Manual, in Appendix C.

This one is a little tricky to answer, but here's what I can give: In the Intel Instruction Set Reference, they are all grouped together under one common instruction, Jcc (Jump if condition is met). I'd like to address the comment that nothing indicates that the different jump instructions take the same amount of time. The two will take the same amount of time. So the only difference between the two is a jg versus a jge instruction.

jg - Jump if greater -> ZF = 0 and SF = OFĮxample (Edited for brevity) Compiled with $ gcc -m32 -S -masm=intel test.c if (a = b.

And a Jcc (jump) instruction, depending on the comparison type (and code layout):.

A test or cmp instruction, which sets EFLAGS.

You didn't specify, but on x86, all of the integral comparisons will be typically implemented in two machine instructions: No, it will not be faster on most architectures.