Wrong evaluation of pre- and post-increment on variables: What is wrong here?!?!
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I tried the following in VS6.0. Created a simple "Hello World" application then added the following:
int i;
int j;
i = j = 5;printf("%d %d %d %d %d %d\n",++i,j++,i++,++j,++i,j++);
printf("i %d j %d\n",i,j);Now i expected the output to be: 8 7 6 7 6 5 i 8 j 8 But instead it printed as: 7 6 6 6 6 5 i 8 j 8 :confused::confused::confused:
Ralph_2 wrote:
Now i expected the output to be: 8 7 6 7 6 5
You cannot expect that behaviour (and you should avoid using the increment operators that way). :)
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
[My articles] -
Ralph_2 wrote:
Now i expected the output to be: 8 7 6 7 6 5
You cannot expect that behaviour (and you should avoid using the increment operators that way). :)
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
[My articles]CPallini wrote:
and you should avoid using the increment operators that way
Oh, this is just for trying...I don't use it any where as such
CPallini wrote:
You cannot expect that behaviour
Why cannot i expect the output to be that? You think the output is wrong? Or different compilers will treat the statement differently??
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CPallini wrote:
and you should avoid using the increment operators that way
Oh, this is just for trying...I don't use it any where as such
CPallini wrote:
You cannot expect that behaviour
Why cannot i expect the output to be that? You think the output is wrong? Or different compilers will treat the statement differently??
Ralph_2 wrote:
Or different compilers will treat the statement differently??
Yes. :)
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
[My articles] -
I tried the following in VS6.0. Created a simple "Hello World" application then added the following:
int i;
int j;
i = j = 5;printf("%d %d %d %d %d %d\n",++i,j++,i++,++j,++i,j++);
printf("i %d j %d\n",i,j);Now i expected the output to be: 8 7 6 7 6 5 i 8 j 8 But instead it printed as: 7 6 6 6 6 5 i 8 j 8 :confused::confused::confused:
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Behavior in this case depends on in which order the operands are pushed on the stack( Right to left or Left to right) when printf is called.
Regards, Sandip.
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Ralph_2 wrote:
Or different compilers will treat the statement differently??
Yes. :)
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
[My articles] -
True. But the Standard Calling Convention for C is right-to-left. The output 7 6 6 6 6 5 does not seem to be correct even if you were to pass the values left to right :)
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I dont see anything wrong with the output? ++i,j++,i++,++j,++i,j++ 1. j++ = 5 2. ++i = 6 3. ++j = 6 4. i++ = 6 5. j++ = 6 7. ++i = 7 You can read more about it here Calling Conventions Demystified[^]
Regards, Sandip.
Your example is wrong! I dont see anything wrong with the output? ++i,j++,i++,++j,++i,j++ 1. j++ = 5 true -> j = 5 but after this statement j=6! 2. ++i = 6 true 3. ++j = 6 wrong -> j=7! 4. i++ = 6 true, but after this statement i=7! 5. j++ = 6 wrong -> j=7! 7. ++i = 7 wrong -> i = 8 So the final result is: 8 7 6 7 6 5 Thats also the right compiler result!
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I dont see anything wrong with the output? ++i,j++,i++,++j,++i,j++ 1. j++ = 5 2. ++i = 6 3. ++j = 6 4. i++ = 6 5. j++ = 6 7. ++i = 7 You can read more about it here Calling Conventions Demystified[^]
Regards, Sandip.
I will surely read the article, i see a lot many interesting points there. But before that, just a little curious, 1. j++ = 5 //OK. Printed value is 5, after that it becomes 6 2. ++i = 6 //OK. First i is incremented to 6 then printed out 3. ++j = 6 //Incorrect.j will first be incremented then printed out. But note that present value of j is 6 (from step 1). So j at this stage becomes 7. 4. i++ = 6 //OK. i now is 7 after printing 5. j++ = 6 //Incorrect. Should have been 7 6. ++i = 7 //Incorrect. Should have been 8 Just to make things a little simpler
i = j = 5;
printf("%d %d %d\n",i++,i++,i++);
printf("i %d\n",i);Now the output is: 5 5 5 i 8 Even if the beahviour is compiler-specific how do you interpret this example (or even the first one)?
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Your example is wrong! I dont see anything wrong with the output? ++i,j++,i++,++j,++i,j++ 1. j++ = 5 true -> j = 5 but after this statement j=6! 2. ++i = 6 true 3. ++j = 6 wrong -> j=7! 4. i++ = 6 true, but after this statement i=7! 5. j++ = 6 wrong -> j=7! 7. ++i = 7 wrong -> i = 8 So the final result is: 8 7 6 7 6 5 Thats also the right compiler result!
I did not see your post before posting mine....i agree with you....the only viable answer so far seems to be that such evaluations are "compiler-specific" but even then I am not able to derive a logic for the output 5 6 6 6 6 7 !!! After all compilers are not like girl-friends :-\ there has to be some logic or rules that determine the output.
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Your example is wrong! I dont see anything wrong with the output? ++i,j++,i++,++j,++i,j++ 1. j++ = 5 true -> j = 5 but after this statement j=6! 2. ++i = 6 true 3. ++j = 6 wrong -> j=7! 4. i++ = 6 true, but after this statement i=7! 5. j++ = 6 wrong -> j=7! 7. ++i = 7 wrong -> i = 8 So the final result is: 8 7 6 7 6 5 Thats also the right compiler result!
Covean wrote:
Your example is wrong! I dont see anything wrong with the output? ++i,j++,i++,++j,++i,j++ 1. j++ = 5 true -> j = 5 but after this statement j=6!
Please Try following sample
int i=5;
printf("%d %d %d %d %d %d",i++,i++,i++,i++,i++,i++);According to you what should be the output? 11 10 9 8 7 6 5? With VS 6.0 the output is 5 5 5 5 5 5 i++ is post increment and will be executed after execution of printf. Please correct me if i am wrong?
Regards, Sandip.
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I will surely read the article, i see a lot many interesting points there. But before that, just a little curious, 1. j++ = 5 //OK. Printed value is 5, after that it becomes 6 2. ++i = 6 //OK. First i is incremented to 6 then printed out 3. ++j = 6 //Incorrect.j will first be incremented then printed out. But note that present value of j is 6 (from step 1). So j at this stage becomes 7. 4. i++ = 6 //OK. i now is 7 after printing 5. j++ = 6 //Incorrect. Should have been 7 6. ++i = 7 //Incorrect. Should have been 8 Just to make things a little simpler
i = j = 5;
printf("%d %d %d\n",i++,i++,i++);
printf("i %d\n",i);Now the output is: 5 5 5 i 8 Even if the beahviour is compiler-specific how do you interpret this example (or even the first one)?
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I will surely read the article, i see a lot many interesting points there. But before that, just a little curious, 1. j++ = 5 //OK. Printed value is 5, after that it becomes 6 2. ++i = 6 //OK. First i is incremented to 6 then printed out 3. ++j = 6 //Incorrect.j will first be incremented then printed out. But note that present value of j is 6 (from step 1). So j at this stage becomes 7. 4. i++ = 6 //OK. i now is 7 after printing 5. j++ = 6 //Incorrect. Should have been 7 6. ++i = 7 //Incorrect. Should have been 8 Just to make things a little simpler
i = j = 5;
printf("%d %d %d\n",i++,i++,i++);
printf("i %d\n",i);Now the output is: 5 5 5 i 8 Even if the beahviour is compiler-specific how do you interpret this example (or even the first one)?
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CPallini wrote:
and you should avoid using the increment operators that way
Oh, this is just for trying...I don't use it any where as such
CPallini wrote:
You cannot expect that behaviour
Why cannot i expect the output to be that? You think the output is wrong? Or different compilers will treat the statement differently??
Ralph_2 wrote:
Or different compilers will treat the statement differently??
Yes. It's all about sequence points.
"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
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Covean wrote:
Your example is wrong! I dont see anything wrong with the output? ++i,j++,i++,++j,++i,j++ 1. j++ = 5 true -> j = 5 but after this statement j=6!
Please Try following sample
int i=5;
printf("%d %d %d %d %d %d",i++,i++,i++,i++,i++,i++);According to you what should be the output? 11 10 9 8 7 6 5? With VS 6.0 the output is 5 5 5 5 5 5 i++ is post increment and will be executed after execution of printf. Please correct me if i am wrong?
Regards, Sandip.
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I tried the following in VS6.0. Created a simple "Hello World" application then added the following:
int i;
int j;
i = j = 5;printf("%d %d %d %d %d %d\n",++i,j++,i++,++j,++i,j++);
printf("i %d j %d\n",i,j);Now i expected the output to be: 8 7 6 7 6 5 i 8 j 8 But instead it printed as: 7 6 6 6 6 5 i 8 j 8 :confused::confused::confused:
All of this "I expected..." and "it should be..." stuff is pointless. Compiler vendors are free to evaluate such expressions as they wish just so long as all side effects of previous evaluations have been performed at each sequence point. In your printf example, the order in which the arguments are evaluated is not specified, but the sequence point means that all of their side effects are complete before
printf()is entered."Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
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All of this "I expected..." and "it should be..." stuff is pointless. Compiler vendors are free to evaluate such expressions as they wish just so long as all side effects of previous evaluations have been performed at each sequence point. In your printf example, the order in which the arguments are evaluated is not specified, but the sequence point means that all of their side effects are complete before
printf()is entered."Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
DavidCrow wrote:
All of this "I expected..." and "it should be..." stuff is pointless
What I mean to imply here is that based on what I have read/know the output should have been so and so.....So may be my understanding of this concept (of calling convention, sequence points) needs to be brushed up....It *could be* my ignorance on the matter, lets see, I'll read more, do more research and see what's missing :)
DavidCrow wrote:
Compiler vendors are free to evaluate such expressions as they wish just so long as all side effects of previous evaluations have been performed at each sequence point.
Expressions....true, but the point here is about "passing variables" and C does specify how to do that: either R-to-L or L-to-R, we can make that selection.
DavidCrow wrote:
In your printf example, the order in which the arguments are evaluated is not specified,
I did not specify that but I m using the default calling convention under which variables are passed R-to-L
