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How do I put two increment statements in a C++ 'for' loop?

Posted by: admin November 29, 2017 Leave a comment

Questions:

I would like to increment two variables in a for-loop condition instead of one.

So something like:

for (int i = 0; i != 5; ++i and ++j) 
    do_something(i, j);

What is the syntax for this?

Answers:

A common idiom is to use the comma operator which evaluates both operands, and returns the second operand. Thus:

for(int i = 0; i != 5; ++i,++j) 
    do_something(i,j);

But is it really a comma operator?

Now having wrote that, a commenter suggested it was actually some special syntactic sugar in the for statement, and not a comma operator at all. I checked that in GCC as follows:

int i=0;
int a=5;
int x=0;

for(i; i<5; x=i++,a++){
    printf("i=%d a=%d x=%d\n",i,a,x);
}

I was expecting x to pick up the original value of a, so it should have displayed 5,6,7.. for x. What I got was this

i=0 a=5 x=0
i=1 a=6 x=0
i=2 a=7 x=1
i=3 a=8 x=2
i=4 a=9 x=3

However, if I bracketed the expression to force the parser into really seeing a comma operator, I get this

int main(){
    int i=0;
    int a=5;
    int x=0;

    for(i=0; i<5; x=(i++,a++)){
        printf("i=%d a=%d x=%d\n",i,a,x);
    }
}

i=0 a=5 x=0
i=1 a=6 x=5
i=2 a=7 x=6
i=3 a=8 x=7
i=4 a=9 x=8

Initially I thought that this showed it wasn’t behaving as a comma operator at all, but as it turns out, this is simply a precedence issue – the comma operator has the lowest possible precedence, so the expression x=i++,a++ is effectively parsed as (x=i++),a++

Thanks for all the comments, it was an interesting learning experience, and I’ve been using C for many years!

Questions:
Answers:

Try this

for(int i = 0; i != 5; ++i, ++j)
    do_something(i,j);

Questions:
Answers:

Try not to do it!

From http://www.research.att.com/~bs/JSF-AV-rules.pdf:

AV Rule 199
The increment expression in a for loop will perform no action other than to change a single
loop parameter to the next value for the loop.

Rationale: Readability.

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Answers:
for (int i = 0; i != 5; ++i, ++j) 
    do_something(i, j);

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I came here to remind myself how to code a second index into the increment clause of a FOR loop, which I knew could be done mainly from observing it in a sample that I incorporated into another project, that written in C++.

Today, I am working in C#, but I felt sure that it would obey the same rules in this regard, since the FOR statement is one of the oldest control structures in all of programming. Thankfully, I had recently spent several days precisely documenting the behavior of a FOR loop in one of my older C programs, and I quickly realized that those studies held lessons that applied to today’s C# problem, in particular to the behavior of the second index variable.

For the unwary, following is a summary of my observations. Everything I saw happening today, by carefully observing variables in the Locals window, confirmed my expectation that a C# FOR statement behaves exactly like a C or C++ FOR statement.

  1. The first time a FOR loop executes, the increment clause (the 3rd of its three) is skipped. In Visual C and C++, the increment is generated as three machine instructions in the middle of the block that implements the loop, so that the initial pass runs the initialization code once only, then jumps over the increment block to execute the termination test. This implements the feature that a FOR loop executes zero or more times, depending on the state of its index and limit variables.
  2. If the body of the loop executes, its last statement is a jump to the first of the three increment instructions that were skipped by the first iteration. After these execute, control falls naturally into the limit test code that implements the middle clause. The outcome of that test determines whether the body of the FOR loop executes, or whether control transfers to the next instruction past the jump at the bottom of its scope.
  3. Since control transfers from the bottom of the FOR loop block to the increment block, the index variable is incremented before the test is executed. Not only does this behavior explain why you must code your limit clauses the way you learned, but it affects any secondary increment that you add, via the comma operator, because it becomes part of the third clause. Hence, it is not changed on the first iteration, but it is on the last iteration, which never executes the body.

If either of your index variables remains in scope when the loop ends, their value will be one higher than the threshold that stops the loop, in the case of the true index variable. Likewise, if, for example, the second variable is initialized to zero before the loop is entered, its value at the end will be the iteration count, assuming that it is an increment (++), not a decrement, and that nothing in the body of the loop changes its value.

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I agree with squelart. Incrementing two variables is bug prone, especially if you only test for one of them.

This is the readable way to do this:

for(int i = 0; i < 5; ++i) {
    ++j;
    do_something(i, j);
}

For loops are meant for cases where your loop runs on one increasing/decreasing variable. For any other variable, change it in the loop.

If you need j to be tied to i, why not leave the original variable as is and add i?

for(int i = 0; i < 5; ++i) {
    do_something(i,a+i);
}

If your logic is more complex (for example, you need to actually monitor more than one variable), I’d use a while loop.

Questions:
Answers:
int main(){
    int i=0;
    int a=0;
    for(i;i<5;i++,a++){
        printf("%d %d\n",a,i);
    } 
}

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Answers:

Use Maths. If the two operations mathematically depend on the loop iteration, why not do the math?

int i, j;//That have some meaningful values in them?
for( int counter = 0; counter < count_max; ++counter )
    do_something (counter+i, counter+j);

Or, more specifically referring to the OP’s example:

for(int i = 0; i != 5; ++i)
    do_something(i, j+i);

Especially if you’re passing into a function by value, then you should get something that does exactly what you want.