Easily understand the essential difference between ++i and i++ from a memory perspective (you can understand it at a glance)

Easily understand the essential difference between ++i and i++ from a memory perspective (you can understand it at a glance)

In this article, I will mainly introduce some of the underlying logic differences between ++i and i++. I believe you have also seen a lot of information on the Internet that states: ++i first increments itself by +1 and then uses it (that is, then performs an expression operation); while i++ is used first and then increments by +1.

Next, we look at this process from a memory perspective through the following code and the assembly of the code.
You don’t need to know much assembly code to understand this.

int main()
{
    
    
    int b = 0;
    int a = 0;
    b = a++ //此时为后置++，即先使用，之后再进行++
   	return 0;
}

The following is the assembly code obtained by disassembly of this code in VS2019. The

key analysis starts from b = a++;
The first mov instruction: represents that the variable a is first assigned to the eax register (the meaning of the register can be regarded as a temporary The role of storage);
the second mov instruction: then the value of the eax register is assigned to the variable b , so b gets this value.
The third mov instruction: at this time the value of the variable a is assigned to the ecx register (note the difference with eax is different above)
The fourth step is the add instruction: it means to +1 the value of ecx . The last step: the move instruction is to return the value after +1 to the variable a , so the value of a is 1 more at this time.

At this point we change a++ into ++a and look at the process again

int main()
{
    
    
	int b = 0;
	int a = 0;
	b = ++a;//前置++,先＋1，再使用
	return 0;
}

Let’s start with b= ++a;.
The first step is the mov instruction: assign the value of the variable a to the eax register.
The second step: the add instruction: add the value of eax + 1 (this step is also the key to distinguishing it from the above).
The third step mov instruction: The value of the eax register is returned to a, which is already the value after +1. The next
two steps are similar to the above. After the value of a is given to ecx, ecx assigns it to variable b.

From the above analysis, we can quickly understand why there is a +1 time difference between ++i and i++.

This is not over yet. Sometimes we have some concerns about ++i or i++ that appear alone. Here we will also conduct a wave of analysis.

int i = 0;
while (i < 32)
{
    
    
	printf("%d", i);
	i++; //此处也可以改成i++
}

If ++ or – appears alone like this, the essence is the same.
The old rule is to look at the assembly code!

Here we only focus on the i++ step.
The first step: mov is to give the value of i to the register first.
The second step: just +1 directly (and in our above code, it has been used before, so the value will be returned to the user and then Perform the ++ operation, but there is no such operation here)
The third step is to return the value after adding 1 to the variable i

Next we change it to ++i;

First of all, the first step is to give i to eax, the second step is to directly +1. The third step returns the value to i.

Summary : So when ++i or i++ appears alone (without a user), they actually mean the same thing.

PS : The last thing I want to tell you is that when you see these assembly codes, you may have this question in your mind, why do you need to use a register like eax every time?
In fact, the reason is very simple. The value of the variable is essentially in the memory , but once these values ​​are calculated, they will be loaded into the CPU . The calculation will only be performed in the CPU . They are first stored in the registers of the CPU and then used by the CPU. arithmetic and logical operations

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