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Difference between i++ and (i)++ in C
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int i = 3;
int j = (i)++;
vs
int i = 3;
int j = i ++;
Though both of the above examples store 3 in j, is there a difference between how the above two cases are evaluated?
Since i is an int, do the () cause the first case to be evaluated as an expression, which would be equivalent to incrementing an rvalue? Or is it undefined behaviour and just happens to store 3 in j?
Or am I overthinking it and its just a simple postfix?
c
edited 5 hours ago
Richard Chambers
10k24167
asked 5 hours ago
Polaris000Polaris000
131412
add a comment |
int i = 3;
int j = (i)++;
vs
int i = 3;
int j = i ++;
Though both of the above examples store 3 in j, is there a difference between how the above two cases are evaluated?
Since i is an int, do the () cause the first case to be evaluated as an expression, which would be equivalent to incrementing an rvalue? Or is it undefined behaviour and just happens to store 3 in j?
Or am I overthinking it and its just a simple postfix?
c
edited 5 hours ago
Richard Chambers
10k24167
asked 5 hours ago
Polaris000Polaris000
131412
2
Seemingly arbitrary usage of parentheses is common in macro definitions. Where they do make a big difference, you'd like the error message you get. Well, usually.
– Hans Passant
4 hours ago
add a comment |
int i = 3;
int j = (i)++;
vs
int i = 3;
int j = i ++;
Though both of the above examples store 3 in j, is there a difference between how the above two cases are evaluated?
Since i is an int, do the () cause the first case to be evaluated as an expression, which would be equivalent to incrementing an rvalue? Or is it undefined behaviour and just happens to store 3 in j?
Or am I overthinking it and its just a simple postfix?
c
edited 5 hours ago
Richard Chambers
10k24167
asked 5 hours ago
Polaris000Polaris000
131412
int i = 3;
int j = (i)++;
vs
int i = 3;
int j = i ++;
Though both of the above examples store 3 in j, is there a difference between how the above two cases are evaluated?
Since i is an int, do the () cause the first case to be evaluated as an expression, which would be equivalent to incrementing an rvalue? Or is it undefined behaviour and just happens to store 3 in j?
Or am I overthinking it and its just a simple postfix?
c
c
edited 5 hours ago
Richard Chambers
10k24167
asked 5 hours ago
Polaris000Polaris000
131412
edited 5 hours ago
Richard Chambers
10k24167
asked 5 hours ago
Polaris000Polaris000
131412
edited 5 hours ago
Richard Chambers
10k24167
edited 5 hours ago
Richard Chambers
10k24167
edited 5 hours ago
Richard Chambers
10k24167
10k24167
asked 5 hours ago
Polaris000Polaris000
131412
asked 5 hours ago
Polaris000Polaris000
131412
asked 5 hours ago
Polaris000Polaris000
131412
131412
2
Seemingly arbitrary usage of parentheses is common in macro definitions. Where they do make a big difference, you'd like the error message you get. Well, usually.
– Hans Passant
4 hours ago
add a comment |
2
Seemingly arbitrary usage of parentheses is common in macro definitions. Where they do make a big difference, you'd like the error message you get. Well, usually.
– Hans Passant
4 hours ago
2
2
Seemingly arbitrary usage of parentheses is common in macro definitions. Where they do make a big difference, you'd like the error message you get. Well, usually.
– Hans Passant
4 hours ago
Seemingly arbitrary usage of parentheses is common in macro definitions. Where they do make a big difference, you'd like the error message you get. Well, usually.
– Hans Passant
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
i++ and (i)++ behave identically. C 2018 6.5.1 5 says:
A parenthesized expression is a primary expression. Its type and value are identical to those of the unparenthesized expression. It is an lvalue, a function designator, or a void expression if the unparenthesized expression is, respectively, an lvalue, a function designator, or a void expression.
answered 5 hours ago
Eric PostpischilEric Postpischil
77.3k881164
add a comment |
In your simple example of i++ versus (i)++, there is no difference, as noted in Eric Postpischil's answer.
However, this difference is actually meaningful if you are dereferencing a pointer variable with the * operator and using the increment operator; there is a difference between *p++ and (*p)++.
The former statement dereferences the pointer and then increments the pointer itself; the latter statement dereferences the pointer then increments the dereferenced value.
answered 5 hours ago
Govind ParmarGovind Parmar
11.8k53361
1
@downvoter: reason?
– Govind Parmar
5 hours ago
1
@SerG In my experience this particular nuance leads to a very common error by beginners, so I thought it was worth mentioning here.
– Govind Parmar
5 hours ago
An attempt to make information available through denormalization instead of structuration is a controversial way.
– SerG
5 hours ago
1
"increments the pointer and then dereferences it" may be confusing, since you're using postincrement -- I would suggest "dereferences the pointer and then increments the pointer".
– zwol
5 hours ago
Your answer: a) acknowledges that question has already been answered; b) goes on a tangent about the case that question didn't mention.
– n0rd
2 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
i++ and (i)++ behave identically. C 2018 6.5.1 5 says:
A parenthesized expression is a primary expression. Its type and value are identical to those of the unparenthesized expression. It is an lvalue, a function designator, or a void expression if the unparenthesized expression is, respectively, an lvalue, a function designator, or a void expression.
answered 5 hours ago
Eric PostpischilEric Postpischil
77.3k881164
add a comment |
i++ and (i)++ behave identically. C 2018 6.5.1 5 says:
A parenthesized expression is a primary expression. Its type and value are identical to those of the unparenthesized expression. It is an lvalue, a function designator, or a void expression if the unparenthesized expression is, respectively, an lvalue, a function designator, or a void expression.
answered 5 hours ago
Eric PostpischilEric Postpischil
77.3k881164
add a comment |
i++ and (i)++ behave identically. C 2018 6.5.1 5 says:
A parenthesized expression is a primary expression. Its type and value are identical to those of the unparenthesized expression. It is an lvalue, a function designator, or a void expression if the unparenthesized expression is, respectively, an lvalue, a function designator, or a void expression.
answered 5 hours ago
Eric PostpischilEric Postpischil
77.3k881164
i++ and (i)++ behave identically. C 2018 6.5.1 5 says:
A parenthesized expression is a primary expression. Its type and value are identical to those of the unparenthesized expression. It is an lvalue, a function designator, or a void expression if the unparenthesized expression is, respectively, an lvalue, a function designator, or a void expression.
answered 5 hours ago
Eric PostpischilEric Postpischil
77.3k881164
answered 5 hours ago
Eric PostpischilEric Postpischil
77.3k881164
answered 5 hours ago
Eric PostpischilEric Postpischil
77.3k881164
answered 5 hours ago
Eric PostpischilEric Postpischil
77.3k881164
77.3k881164
add a comment |
add a comment |
In your simple example of i++ versus (i)++, there is no difference, as noted in Eric Postpischil's answer.
However, this difference is actually meaningful if you are dereferencing a pointer variable with the * operator and using the increment operator; there is a difference between *p++ and (*p)++.
The former statement dereferences the pointer and then increments the pointer itself; the latter statement dereferences the pointer then increments the dereferenced value.
answered 5 hours ago
Govind ParmarGovind Parmar
11.8k53361
1
@downvoter: reason?
– Govind Parmar
5 hours ago
1
@SerG In my experience this particular nuance leads to a very common error by beginners, so I thought it was worth mentioning here.
– Govind Parmar
5 hours ago
An attempt to make information available through denormalization instead of structuration is a controversial way.
– SerG
5 hours ago
1
"increments the pointer and then dereferences it" may be confusing, since you're using postincrement -- I would suggest "dereferences the pointer and then increments the pointer".
– zwol
5 hours ago
Your answer: a) acknowledges that question has already been answered; b) goes on a tangent about the case that question didn't mention.
– n0rd
2 mins ago
add a comment |
In your simple example of i++ versus (i)++, there is no difference, as noted in Eric Postpischil's answer.
However, this difference is actually meaningful if you are dereferencing a pointer variable with the * operator and using the increment operator; there is a difference between *p++ and (*p)++.
The former statement dereferences the pointer and then increments the pointer itself; the latter statement dereferences the pointer then increments the dereferenced value.
answered 5 hours ago
Govind ParmarGovind Parmar
11.8k53361
1
@downvoter: reason?
– Govind Parmar
5 hours ago
1
@SerG In my experience this particular nuance leads to a very common error by beginners, so I thought it was worth mentioning here.
– Govind Parmar
5 hours ago
An attempt to make information available through denormalization instead of structuration is a controversial way.
– SerG
5 hours ago
1
"increments the pointer and then dereferences it" may be confusing, since you're using postincrement -- I would suggest "dereferences the pointer and then increments the pointer".
– zwol
5 hours ago
Your answer: a) acknowledges that question has already been answered; b) goes on a tangent about the case that question didn't mention.
– n0rd
2 mins ago
add a comment |
In your simple example of i++ versus (i)++, there is no difference, as noted in Eric Postpischil's answer.
However, this difference is actually meaningful if you are dereferencing a pointer variable with the * operator and using the increment operator; there is a difference between *p++ and (*p)++.
The former statement dereferences the pointer and then increments the pointer itself; the latter statement dereferences the pointer then increments the dereferenced value.
answered 5 hours ago
Govind ParmarGovind Parmar
11.8k53361
In your simple example of i++ versus (i)++, there is no difference, as noted in Eric Postpischil's answer.
However, this difference is actually meaningful if you are dereferencing a pointer variable with the * operator and using the increment operator; there is a difference between *p++ and (*p)++.
The former statement dereferences the pointer and then increments the pointer itself; the latter statement dereferences the pointer then increments the dereferenced value.
answered 5 hours ago
Govind ParmarGovind Parmar
11.8k53361
edited 4 hours ago
answered 5 hours ago
Govind ParmarGovind Parmar
11.8k53361
answered 5 hours ago
Govind ParmarGovind Parmar
11.8k53361
answered 5 hours ago
Govind ParmarGovind Parmar
11.8k53361
11.8k53361
1
@downvoter: reason?
– Govind Parmar
5 hours ago
1
@SerG In my experience this particular nuance leads to a very common error by beginners, so I thought it was worth mentioning here.
– Govind Parmar
5 hours ago
An attempt to make information available through denormalization instead of structuration is a controversial way.
– SerG
5 hours ago
1
"increments the pointer and then dereferences it" may be confusing, since you're using postincrement -- I would suggest "dereferences the pointer and then increments the pointer".
– zwol
5 hours ago
Your answer: a) acknowledges that question has already been answered; b) goes on a tangent about the case that question didn't mention.
– n0rd
2 mins ago
add a comment |
1
@downvoter: reason?
– Govind Parmar
5 hours ago
1
@SerG In my experience this particular nuance leads to a very common error by beginners, so I thought it was worth mentioning here.
– Govind Parmar
5 hours ago
An attempt to make information available through denormalization instead of structuration is a controversial way.
– SerG
5 hours ago
1
"increments the pointer and then dereferences it" may be confusing, since you're using postincrement -- I would suggest "dereferences the pointer and then increments the pointer".
– zwol
5 hours ago
Your answer: a) acknowledges that question has already been answered; b) goes on a tangent about the case that question didn't mention.
– n0rd
2 mins ago
1
1
@downvoter: reason?
– Govind Parmar
5 hours ago
@downvoter: reason?
– Govind Parmar
5 hours ago
1
1
@SerG In my experience this particular nuance leads to a very common error by beginners, so I thought it was worth mentioning here.
– Govind Parmar
5 hours ago
@SerG In my experience this particular nuance leads to a very common error by beginners, so I thought it was worth mentioning here.
– Govind Parmar
5 hours ago
An attempt to make information available through denormalization instead of structuration is a controversial way.
– SerG
5 hours ago
An attempt to make information available through denormalization instead of structuration is a controversial way.
– SerG
5 hours ago
1
1
"increments the pointer and then dereferences it" may be confusing, since you're using postincrement -- I would suggest "dereferences the pointer and then increments the pointer".
– zwol
5 hours ago
"increments the pointer and then dereferences it" may be confusing, since you're using postincrement -- I would suggest "dereferences the pointer and then increments the pointer".
– zwol
5 hours ago
Your answer: a) acknowledges that question has already been answered; b) goes on a tangent about the case that question didn't mention.
– n0rd
2 mins ago
Your answer: a) acknowledges that question has already been answered; b) goes on a tangent about the case that question didn't mention.
– n0rd
2 mins ago
add a comment |
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2
Seemingly arbitrary usage of parentheses is common in macro definitions. Where they do make a big difference, you'd like the error message you get. Well, usually.
– Hans Passant
4 hours ago