C语言 从任何角度来看，++i 和 i+=1 之间有什么区别
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what is difference between ++i and i+=1 from any point of view
提问by chanzerre
This is a question from kn king's c programming : a modern approach. I can't understand the solution given by him:-
这是来自 kn king's c programming 的一个问题：一种现代方法。我无法理解他给出的解决方案：-
The expression ++i is equivalent to (i += 1). The value of both expressions is i after
the increment has been performed.
How do I understand this anyway?
无论如何，我如何理解这一点？
回答by thefourtheye
i = 10
printf("%d", i++);
will print 10, where as
将打印 10，其中
printf("%d", ++i);
will print 11
将打印 11
X = i++can be thought as this
X = i++可以这样认为
X = i
i = i + 1
where as X = ++iis
其中，作为X = ++i为
i = i + 1
X = i
so,
所以，
printf ("%d", ++i);
is same as
与
printf ("%d", i += 1);
but not
但不是
printf ("%d", i++);
although value of iafter any of these three statements will be the same.
尽管i这三个语句中任何一个之后的值都是相同的。
回答by user4815162342
The solution means to say that there isno difference, ++ihas the same meaning as (i += 1)no matter what ihappens to be and no matter the context of the expression. The parentheses around i += 1make sure that the equivalence holds even when the context contains further arithmetics, such as ++i * 3being equivalent to (i += 1) * 3, but not to i += 1 * 3(which is equivalent to i += 3).
该解决方案的意思是说，有是没有什么区别，++i具有相同的含义(i += 1)不管什么i恰好是并不管表达的情况下。周围的括号i += 1确保即使上下文包含进一步的算术，例如++i * 3等效于(i += 1) * 3，但不i += 1 * 3等效于（等效于i += 3），等价也成立。
The same would not apply to i++, which has the same side effect (incrementing i), but a different value in the surrounding expression — the value of ibefore being incremented.
这同样不适用于i++，它具有相同的副作用（递增i），但周围表达式中的值不同 -i递增之前的值。
回答by user1952500
One difference that has not been brought up so far is readability of code. A large part of loops use increment by one and common practice is to use i++/++i when moving to the next element / incrementing an index by 1.
到目前为止还没有提出的一个区别是代码的可读性。大部分循环使用 1 递增，通常的做法是在移动到下一个元素/将索引递增 1 时使用 i++/++i。
Typically i+= is used in these cases only when the increment is something other than 1. Using this for the normal increment will not be dangerous but cause a slight bump in the understanding and make the code look unusual.
通常，仅当增量不是 1 时才在这些情况下使用 i+=。将它用于正常增量不会有危险，但会导致理解上的轻微颠簸并使代码看起来不寻常。
回答by naveenKumar
Difference between both are:++ is a unary operator while + is a binary operator....
If we consider execution time:i++ is more faster than i=i+1.No of machine cycles differs to execute the same set of code, thats the reason ++ operators are always prefered for Loops.
Refer to this thread for more info
两者之间的区别是：++ 是一元运算符，而 + 是二元运算符……
如果我们考虑执行时间：i++ is more faster than i=i+1.执行同一组代码的机器周期数不同，这就是 ++ 运算符总是首选的原因循环。
有关更多信息，请参阅此线程
回答by Mike
I think they are totally the same. There is one thing maybe interesting. The ++i is equal to (i+=1) but not i+=1; The difference is the brace. Because i += 1 may depend on the context and it will have different interpretation.
我认为他们是完全一样的。有一件事可能很有趣。++i 等于 (i+=1) 但不等于 i+=1；不同之处在于支架。因为 i += 1 可能取决于上下文，它会有不同的解释。
回答by carlos
In a normal operation without assignation:
在没有赋值的正常操作中：
++i and i++
++我和我++
increase the variable in 1. In pseudo assembly both code it is:
在 1 中增加变量。在伪汇编中，这两个代码都是：
inc i
but If you assign the value the order of ++ is relevant:
但是，如果您分配该值，则 ++ 的顺序是相关的：
x = i++
x = i++
produce:
生产：
mov x, i
inc i
x = ++i
x = ++i
produce:
生产：
inc i
mov x, i
In the case of: i += 1
在以下情况下： i += 1
it will produce:
它将产生：
add i,1
but because compilers optimize the code it also will produce in this case:
但是因为编译器优化了代码，它也会在这种情况下产生：
inc i
回答by phaazon
++iis the pre-increment operator. It increments ibefore setting and returning the value (which is obviously i + 1).
++i是预增量运算符。它i在设置和返回值之前递增（显然是i + 1）。
Now, i++is the post-increment operator. It increments iafter the whole instruction it appears in is evaluated.
现在，i++是后增量运算符。它i在它出现的整个指令被评估后递增。
Example:
例子：
int i = 0;
std::cout << ++i << std::endl; /* you get 1 here */
std::cout << i++ << std::endl; /* you still get 1 here */
std::cout << i << std::endl; /* you get 2 here */
