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Difference between i++ and (i)++ in C

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Difference between i++ and (i)++ in C

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6

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

share|improve this question

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 | 

6

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

share|improve this question

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

share|improve this question

edited 5 hours ago

Richard Chambers

10k24167

asked 5 hours ago

Polaris000Polaris000

131412

int i = 3;

int j = (i)++;

int i = 3;

int j = i ++;

share|improve this question

edited 5 hours ago

Richard Chambers

10k24167

asked 5 hours ago

Polaris000Polaris000

131412

share|improve this question

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

asked 5 hours ago

Polaris000Polaris000

131412

asked 5 hours ago

Polaris000Polaris000

131412

2 Answers
2

active

oldest

votes

20

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.

share|improve this answer

answered 5 hours ago

Eric PostpischilEric Postpischil

77.3k881164

add a comment | 

3

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.

share|improve this answer

edited 4 hours ago

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
  

  
  
  
  

add a comment | 

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20

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.

share|improve this answer

answered 5 hours ago

Eric PostpischilEric Postpischil

77.3k881164

add a comment | 

20

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.

share|improve this answer

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.

share|improve this answer

answered 5 hours ago

Eric PostpischilEric Postpischil

77.3k881164

share|improve this answer

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

3

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.

share|improve this answer

edited 4 hours ago

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
  

  
  
  
  

add a comment | 

3

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.

share|improve this answer

edited 4 hours ago

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
  

  
  
  
  

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.

share|improve this answer

edited 4 hours ago

answered 5 hours ago

Govind ParmarGovind Parmar

11.8k53361

share|improve this answer

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