How does the C expression s[i ++] = t[j ++] exactly work in order?

Question

I'm new in C, stuck in understanding the expression s[i ++] = t[j ++], I don't know how it's possible that an array element gets accessed with a variable and then the variable increase itself and then the array element just accessed is again accessed with the original variable and then gets assigned to another array's element, I'm confused, I think to understand the exact process might involve some low-level knowledges, but I don't want to digress too far away, is there any way to understand it easily and clearly?

Answer 1

In C language, the expression i++ causes i to be incremented, but the expression i++ itself evaluates to the value i had before being incremented. So the expression s[i++] = t[j++] has the same behaviour as:

s[i] = t[j];
i = i + 1;
j = j + 1;

except that the precise order is not specified. For that last reason, the rule is that a variable should only be modified once: s[i++] = t[i++] would invoke Undefined Behaviour.

Answer 2

Like any other complicated-looking expression, it's easier to understand this if you break it down into parts.

The key is that innermost part (or "subexpression") i++. I assume you know what i++ does by itself, although in this example, we're hopefully going to get a deeper appreciation of what i++ is actually good for. Why would you want to "increment i, but return the old value"? What's the use of this? Well, the main use is that it's super useful for moving along an array.

Lets look at a simpler example. Suppose we have an array a that we want to store some numbers in. The most basic way is

int a[10];
a[0] = 12;
a[1] = 34;
a[2] = 5678;

Another very good way is to use a second variable like i to keep track of where we're storing:

i = 0;
a[i] = 12;
i = i + 1;
a[i] = 34;
i = i + 1;
a[i] = 5678;
i = i + 1;

I've written this out in "longhand", but of course in C, you would almost never write it this way, because the "C way" is the much more concise

i = 0;
a[i++] = 12;
a[i++] = 34;
a[i++] = 5678;

So first, make sure you understand that the "shorthand" and "longhand" forms work exactly the same way. Make sure you understand that when we say something like

a[i++] = 34;

what this means is "store 34 into the slot in array a indicated by i, and then update i to be one more than is used to be, so that it indicates the next slot."

In other words, we use an expression like a[i++] whenever we want to move along an array and do something with its elements, one by one, in order.

So far we were storing values into the array, but the idiom works just as well for fetching values out of an array. For example, this code prints those three elements, again one at a time, in order:

i = 0;
printf("%d\n", a[i++]);
printf("%d\n", a[i++]);
printf("%d\n", a[i++]);

My point is, again, that any time you see an expression like a[i++], you should think "we're moving along the array".

So now, finally, we can look at the expression you initially asked about:

s[i++] = t[j++];

Here we have two instances of the idiom. We're using i to move along the array s, and we're using j to move along the array t. We're fetching from t as we move along, and we're storing the values into s.

I don't know whether s and t are arrays of characters, or integers, or what. Also I don't know that s and t are truly arrays -- they might actually be pointers, pointing into some arrays. But I don't really have to know those things to know that the essential meaning of s[i++] = t[j++] is "copy elements from array t to array s, using j to keep where we are in t, and i to keep track of where we are in s".

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[The above is an answer to your original question. The rest of this answer isn't directly related, but is essential to avoid inadvertently writing incorrect programs using ++ and --.]

As I said, the subexpression i++ and the idiom a[i++] are super useful for moving through arrays. But there are a couple things to beware of. (Actually it's just one thing, but it crops up in lots of different ways.)

Earlier I wrote the code

i = 0;
printf("%d\n", a[i++]);
printf("%d\n", a[i++]);
printf("%d\n", a[i++]);

to print the first three elements of the array a. But it prints them as bare, isolated numbers. What if I want to always see which array index each number comes from? That is, what if I'm tempted to write something like this:

i = 0;
printf("%d: %d\n", i, a[i++]);      /* WRONG */
printf("%d: %d\n", i, a[i++]);      /* WRONG */
printf("%d: %d\n", i, a[i++]);      /* WRONG */

If I wrote this, my intent would be that I would see the obvious display

0: 12
1: 34
2: 56788

But when I actually tried it just now, I got this instead:

1: 12
2: 34
3: 5678

The numbers 12 and 34 and 5678 are right, but the indices 1, 2, and 3 are all wrong -- they're off by one! How did that happen?

And the answer is that although i++ is, as I said, "super useful", it turns out that there's a fine line between "super useful" and what's called undefined behavior.

That printf call

printf("%d: %d\n", i, a[i++]);      /* WRONG */

looks fine, but it's not actually well-defined, because the compiler does not necessarily evaluate everything left-to-right, so it's not actually guaranteed that it will use the old value of i for the %d: part. The compiler might evaluate things from right to left, meaning that a[i++] will happen first, meaning that %d: will print the new value, instead -- which appears to be what happened when I tried it.

Here's another potential issue. Your original question was about

s[i++] = t[j++];

which, as we've seen, copies elements from t to s based on two possibly-different indices i and j. But what if we know we always want to copy t[1] to s[1], t[2] to s[2], t[3] to s[3], etc.? That is, what if we know that i and j will always be the same, so we don't even need separate i and j variables? How would we write that? Our first try might be

s[i++] = t[i++];                    /* WRONG */

but that can't be right, because now we're incrementing i twice, and we'll probably do something totally broken like copying t[1] to s[2] and t[3] to s[4]. But if we want to only increment i once, should it be

s[i++] = t[i];                      /* WRONG */

or

s[i] = t[i++];                      /* WRONG */

But the answer is that neither of these will work. In expressions like these, which have i in one place and i++ in the other place, there's no way to tell whether i gets the old value or the new value. (In particular, there's no left-to-right or right-to-left rule that would tell us.)

So although expressions like i++ and a[i++] are indeed super useful, you have to be careful when you use them, to make sure you don't go over the edge and have too much happening at once, such that the evaluation order becomes undefined. Sometimes this means you have to back off, and not use the "super useful" idiom, after all. For example, a safe way to print those values would be

printf("%d: %d\n", i, a[i]); i++;
printf("%d: %d\n", i, a[i]); i++;
printf("%d: %d\n", i, a[i]); i++;

and a safe way to copy from t[1] to s[i] would be

s[i] = t[i]; i++;

You can read more in this answer about how to recognize well-defined expressions involving ++ and --, and how to avoid the undefined ones.

Answer 3

The evaluations of s[i++] and t[j++] are unsequenced relative to each other. Semantically, it's equivalent to:

t1 = i;
t2 = j;

s[t1] = t[t2]; 
i = i + 1;
j = j + 1;

with the caveat that the last three assignments can happen in any order, even simultaneously (either in parallel or interleaved)¹. The compiler doesn't have to create temporaries, either - the whole thing can be evaluated as

s[i] = t[j];
i = i + 1;
j = j + 1;

Alternately, the side effects of i++ and j++ can be applied before the update to s:

t1 = j;
j = j + 1;
t2 = i;
i = i + 1;
s[t2] = t[t1];

The current values of i and j must be known before you can index into the arrays, and the value of t[j] must be known before it can be assigned to s[i], but beyond that there's no fixed order of evaluation or of the application of side effects.

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1. This is why expressions like x = x++ or a = b++ * b++ or a[i] = i++ all invoke undefined behavior - there's no fixed order for evaluating or applying side effects, so the results can vary by compiler, compiler settings, even by the surrounding code, and the results don't have to be consistent from run to run.