Discuss a C language problem

mpc_bywind

Everyone knows the order of function calls in C, but I don't quite understand this problem.
printf ("% d,% d", i, i ++);
printf ("% d,% d", i, ++ i);
printf ("% d,% d", i ++, i);
printf ("% d,% d", ++ i, i);
It's elusive. The order is inconsistent ~~~

Master tie up ~~~~

wanxiaohai

It won't be clear if you run it on the machine

mpc_bywind

It is because the order of running on the computer feels inconsistent. I don't know what is going on.

babyblock

The difference between i ++ and ++ i ~~~~
i ++ increases by 1 after i use ++ i increases by 1 before i use
This way you can see the process more clearly ~ ...

darkhanyi

You just need to know the difference between i ++ and ++ i
As for whether it is from left to right or right to left, you don't need to care.
If you find that the result on one compiler is different from the result on another compiler, you don't need to worry, this is normal.

mpc_bywind

No, no, there will be different orders in a compiler. I will give the results as you know, suppose i = 2
printf ("% d,% d", i, i ++); // This case is displayed as: 3,2
printf ("% d,% d", i, ++ i); // This case is displayed as: 3,3
printf ("% d,% d", i ++, i); // This case is displayed as: 2,3
printf ("% d,% d", ++ i, i); // This case is displayed as: 3,3

Obviously, the order of parameter invocation of the two statements is opposite. How can this happen in a compiler?

魍魉vs琵琶

Do n’t drill into such a problem, know the principle
When running,
I write these programs as library functions and fall out when needed.
Hehe

xufeng001

The direction of parameter passing in C is different from Fortan. It will be learned in the compilation principle. The function call test (a, b), the parameter passing order is actually b, a, (Fortan is a, b)
i = 0;
printf ("% d,% d", i, i ++);
Incoming is i ++: 0, i: 1
Printing in sequence or i: i ++, 1: 0

zjwzjw

Such an execution statement,
The order of execution is not specified in the standard,
Then, it will be decided by the compiler itself.

and so,
The execution result of such a statement depends on the compiler.

As for "opposite",
It's just that your understanding is inconsistent with the implementation of the compiler ...

zjwzjw

printf ("% d,% d\n", i, i ++); // This case is displayed as: 3,2
    printf ("% d,% d\n", i, ++ i); // This case is displayed as: 3,3
    printf ("% d,% d\n", i ++, i); // This case is displayed as: 2,2
    printf ("% d,% d\n", ++ i, i); // This case is displayed as: 3,2

This result may be easier to understand ～～

As for the compiler you use,
Why did you get that result,
Need to analyze compiler behavior,
This is not necessary,
Because analysis is useless,
The feeling is to find a reasonable "excuse" to get a certain result,
Doesn't make any sense ...