How does the compiler evaluate a condition in C

问题

I have a question regarding how the compiler evaluates 'AND' condition in c.

Say, I write a statement like

if( (today is Thursday) && (Month is July) )
{
 //do something
}

Assume today is not Thursday, but the Month is indeed July.

Does the compiler check both the conditions, and do 1&0 == 0? Or as soon as it sees that today is not Thursday, it just skips out and doesn't even bother to check the Month condition since it's inconsequential.

I'm using the following gcc

Using built-in specs.
COLLECT_GCC=gcc
COLLECT_LTO_WRAPPER=/usr/lib/gcc/x86_64-linux-gnu/4.6/lto-wrapper
Target: x86_64-linux-gnu
Configured with: ../src/configure -v --with-pkgversion='Ubuntu/Linaro 4.6.3-1ubuntu5' --with-bugurl=file:///usr/share/doc/gcc-4.6/README.Bugs --enable-languages=c,c++,fortran,objc,obj-c++ --prefix=/usr --program-suffix=-4.6 --enable-shared --enable-linker-build-id --with-system-zlib --libexecdir=/usr/lib --without-included-gettext --enable-threads=posix --with-gxx-include-dir=/usr/include/c++/4.6 --libdir=/usr/lib --enable-nls --with-sysroot=/ --enable-clocale=gnu --enable-libstdcxx-debug --enable-libstdcxx-time=yes --enable-gnu-unique-object --enable-plugin --enable-objc-gc --disable-werror --with-arch-32=i686 --with-tune=generic --enable-checking=release --build=x86_64-linux-gnu --host=x86_64-linux-gnu --target=x86_64-linux-gnu
Thread model: posix
gcc version 4.6.3 (Ubuntu/Linaro 4.6.3-1ubuntu5)

回答1:

According to the C Standard (6.5.13 Logical AND operator)

4 Unlike the bitwise binary & operator, the && operator guarantees left-to-right evaluation; if the second operand is evaluated, there is a sequence point between the evaluations of the first and second operands. If the first operand compares equal to 0, the second operand is not evaluated.

As for the logical OR operator (6.5.14 Logical OR operator) then

4 Unlike the bitwise | operator, the || operator guarantees left-to-right evaluation; if the second operand is evaluated, there is a sequence point between the evaluations of the first and second operands. If the first operand compares unequal to 0, the second operand is not evaluated.

回答2:

As others have already said && is a short circuit operator which in case of && will terminate evaluation as soon as the result is known (which happens when going L-> any operand evaluates to to zero/false which is sufficient condition to cease further evaluation since 0 && a = 0)

Shortly your pseudo-code:

if( (today is Thursday) && (Month is July) )
{
 //do something
}

is equivalent to this:

if (today is Thursday)
{
    if (Month is July)
    {
        //do something
    }
}

回答3:

The && operator short circuits under C. This means if the first condition evaluated fails and it is enough to decide on the result of the expression it will terminate there.

回答4:

using this chart:

http://www.swansontec.com/sopc.html

you will see that the && operator is evaluated from left to right.

Once a 'false' expression is encountered in a &&, the execution 'shortcuts' out of the rest of the evaluation.

So if the first expression is false, then the second expression is not evaluated

来源：https://stackoverflow.com/questions/31436721/how-does-the-compiler-evaluate-a-condition-in-c

标签

logical-operators

short-circuiting