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Why does pointer to an address of a double variable not work in C? [duplicate]

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1

1

This question is an exact duplicate of:

• 
  c-behaviour of printf(“%d”, <double>) [duplicate]
  
  1 answer
  
  

following the example : (working with x64 cpu and x64 GCC)

int a=7; printf("%d",*&a); //It prints 7
double a=7; printf("%d",*&a); //It prints an address

edited: thanks, i got it, i have to use %lf, but since 7 is stored within 1 byte , now i just want to access that byte. so, how to point to that lower byte of the double variable?

c printf double

share|improve this question

edited Mar 22 at 7:10

bipul kalita

asked Mar 22 at 6:48

bipul kalitabipul kalita

197

marked as duplicate by Antti Haapala c
Users with the  c badge can single-handedly close c questions as duplicates and reopen them as needed.

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Mar 22 at 7:01

This question was marked as an exact duplicate of an existing question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You don't assume that the result 7 is the address of a, do you? The expression *& takes an address and then dereferences it. Remove the * to actually get the address.
  
  – harper
  Mar 22 at 6:50
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Use %lf to print double variable not %d. And then this *&a doesn't print address as both * and & nullify each other and it prints 7.000000.
  
  – Achal
  Mar 22 at 6:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Removing the '*' will result in undefined behavior on x64 as typesize for a pointer exceeds storage for int.
  
  – David C. Rankin
  Mar 22 at 6:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Your edit made the question worse. No, there is no "byte with 7"
  
  – Antti Haapala
  Mar 22 at 7:20
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @bipulkalita I repeat. There is no byte with value 7, no matter how much you try to argue the case. So there is no way you're going to find a byte with value seven, period. Please google for IEEE 754
  
  – Antti Haapala
  Mar 22 at 7:24
  

  
  
  
  

 | 
show 2 more comments

1

1

This question is an exact duplicate of:

• 
  c-behaviour of printf(“%d”, <double>) [duplicate]
  
  1 answer
  
  

following the example : (working with x64 cpu and x64 GCC)

int a=7; printf("%d",*&a); //It prints 7
double a=7; printf("%d",*&a); //It prints an address

edited: thanks, i got it, i have to use %lf, but since 7 is stored within 1 byte , now i just want to access that byte. so, how to point to that lower byte of the double variable?

c printf double

share|improve this question

edited Mar 22 at 7:10

bipul kalita

asked Mar 22 at 6:48

bipul kalitabipul kalita

197

marked as duplicate by Antti Haapala c
Users with the  c badge can single-handedly close c questions as duplicates and reopen them as needed.

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Mar 22 at 7:01

This question was marked as an exact duplicate of an existing question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You don't assume that the result 7 is the address of a, do you? The expression *& takes an address and then dereferences it. Remove the * to actually get the address.
  
  – harper
  Mar 22 at 6:50
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Use %lf to print double variable not %d. And then this *&a doesn't print address as both * and & nullify each other and it prints 7.000000.
  
  – Achal
  Mar 22 at 6:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Removing the '*' will result in undefined behavior on x64 as typesize for a pointer exceeds storage for int.
  
  – David C. Rankin
  Mar 22 at 6:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Your edit made the question worse. No, there is no "byte with 7"
  
  – Antti Haapala
  Mar 22 at 7:20
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  @bipulkalita I repeat. There is no byte with value 7, no matter how much you try to argue the case. So there is no way you're going to find a byte with value seven, period. Please google for IEEE 754
  
  – Antti Haapala
  Mar 22 at 7:24
  

  
  
  
  

 | 
show 2 more comments

This question is an exact duplicate of:

• 
  c-behaviour of printf(“%d”, <double>) [duplicate]
  
  1 answer
  
  

following the example : (working with x64 cpu and x64 GCC)

int a=7; printf("%d",*&a); //It prints 7
double a=7; printf("%d",*&a); //It prints an address

edited: thanks, i got it, i have to use %lf, but since 7 is stored within 1 byte , now i just want to access that byte. so, how to point to that lower byte of the double variable?

c printf double

share|improve this question

edited Mar 22 at 7:10

bipul kalita

asked Mar 22 at 6:48

bipul kalitabipul kalita

197

int a=7; printf("%d",*&a); //It prints 7
double a=7; printf("%d",*&a); //It prints an address

share|improve this question

edited Mar 22 at 7:10

bipul kalita

asked Mar 22 at 6:48

bipul kalitabipul kalita

197

share|improve this question

edited Mar 22 at 7:10

bipul kalita

asked Mar 22 at 6:48

bipul kalitabipul kalita

197

asked Mar 22 at 6:48

bipul kalitabipul kalita

197

asked Mar 22 at 6:48

bipul kalitabipul kalita

197

3 Answers
3

active

oldest

votes

3

You don't print a double with %d format specifier. You must use %f.

Using wrong format specifier invokes undefined behaviour. The output (if you get one), cannot be justified in any way.

share|improve this answer

answered Mar 22 at 6:52

Sourav GhoshSourav Ghosh

111k15132191

add a comment | 

1

Because you are printing double with "%d". Change it to "%lf"

Regarding "//It prints an address"

It is not printing address. Printf is taking address of double variable and dereferencing it as an integer. So whatever value it is printing is some junk number.

share|improve this answer

answered Mar 22 at 6:57

MayurKMayurK

1,441619

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  for printf, %lf is a myth. :)
  
  – Sourav Ghosh
  Mar 22 at 7:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @SouravGhosh Just for my understanding: Then why/when "%lf" is required? How printf() differentiates float and double?
  
  – MayurK
  Mar 22 at 8:52
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Actually @MayurK, A float when passed to a printf, will be promoted to double before printf receives it. hence it works with both %f and %lf.
  
  – Rai
  Mar 22 at 8:57
  

  
  
  
  

add a comment | 

1

"Why does pointer to an address of a double variable not work in C?"

No. It works.

double a=7; printf("%d",*&a); //It prints an address

• It's just printing the data at the address specified by &a. The * is for dereferencing the address.

And if you are saying that it's not working because it isn't printing the value of double a, it's simply because you are not using the right format specifier.

• 
  

  %d is the format specifier for int. If you want to print a double use %lf or %f.

  
  
  

  double a=7; printf("%lf",*&a); //This will print the double value

  
  

share|improve this answer

edited Mar 22 at 8:51

answered Mar 22 at 6:59

RaiRai

1,1424823

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  No. It works...that's an overstatement.
  
  – Sourav Ghosh
  Mar 22 at 7:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sorry @SouravGhosh, I didn't quite get you.
  
  – Rai
  Mar 22 at 8:52
  

  
  
  
  

add a comment | 

3

You don't print a double with %d format specifier. You must use %f.

Using wrong format specifier invokes undefined behaviour. The output (if you get one), cannot be justified in any way.

share|improve this answer

answered Mar 22 at 6:52

Sourav GhoshSourav Ghosh

111k15132191

add a comment | 

3

You don't print a double with %d format specifier. You must use %f.

Using wrong format specifier invokes undefined behaviour. The output (if you get one), cannot be justified in any way.

share|improve this answer

answered Mar 22 at 6:52

Sourav GhoshSourav Ghosh

111k15132191

add a comment | 

You don't print a double with %d format specifier. You must use %f.

Using wrong format specifier invokes undefined behaviour. The output (if you get one), cannot be justified in any way.

share|improve this answer

answered Mar 22 at 6:52

Sourav GhoshSourav Ghosh

111k15132191

share|improve this answer

answered Mar 22 at 6:52

Sourav GhoshSourav Ghosh

111k15132191

answered Mar 22 at 6:52

Sourav GhoshSourav Ghosh

111k15132191

answered Mar 22 at 6:52

Sourav GhoshSourav Ghosh

111k15132191

1

Because you are printing double with "%d". Change it to "%lf"

Regarding "//It prints an address"

It is not printing address. Printf is taking address of double variable and dereferencing it as an integer. So whatever value it is printing is some junk number.

share|improve this answer

answered Mar 22 at 6:57

MayurKMayurK

1,441619

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  for printf, %lf is a myth. :)
  
  – Sourav Ghosh
  Mar 22 at 7:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @SouravGhosh Just for my understanding: Then why/when "%lf" is required? How printf() differentiates float and double?
  
  – MayurK
  Mar 22 at 8:52
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Actually @MayurK, A float when passed to a printf, will be promoted to double before printf receives it. hence it works with both %f and %lf.
  
  – Rai
  Mar 22 at 8:57
  

  
  
  
  

add a comment | 

1

Because you are printing double with "%d". Change it to "%lf"

Regarding "//It prints an address"

It is not printing address. Printf is taking address of double variable and dereferencing it as an integer. So whatever value it is printing is some junk number.

share|improve this answer

answered Mar 22 at 6:57

MayurKMayurK

1,441619

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  for printf, %lf is a myth. :)
  
  – Sourav Ghosh
  Mar 22 at 7:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @SouravGhosh Just for my understanding: Then why/when "%lf" is required? How printf() differentiates float and double?
  
  – MayurK
  Mar 22 at 8:52
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Actually @MayurK, A float when passed to a printf, will be promoted to double before printf receives it. hence it works with both %f and %lf.
  
  – Rai
  Mar 22 at 8:57
  

  
  
  
  

add a comment | 

Because you are printing double with "%d". Change it to "%lf"

Regarding "//It prints an address"

It is not printing address. Printf is taking address of double variable and dereferencing it as an integer. So whatever value it is printing is some junk number.

share|improve this answer

answered Mar 22 at 6:57

MayurKMayurK

1,441619

share|improve this answer

answered Mar 22 at 6:57

MayurKMayurK

1,441619

answered Mar 22 at 6:57

MayurKMayurK

1,441619

answered Mar 22 at 6:57

MayurKMayurK

1,441619

1

"Why does pointer to an address of a double variable not work in C?"

No. It works.

double a=7; printf("%d",*&a); //It prints an address

• It's just printing the data at the address specified by &a. The * is for dereferencing the address.

And if you are saying that it's not working because it isn't printing the value of double a, it's simply because you are not using the right format specifier.

• 
  

  %d is the format specifier for int. If you want to print a double use %lf or %f.

  
  
  

  double a=7; printf("%lf",*&a); //This will print the double value

  
  

share|improve this answer

edited Mar 22 at 8:51

answered Mar 22 at 6:59

RaiRai

1,1424823

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  No. It works...that's an overstatement.
  
  – Sourav Ghosh
  Mar 22 at 7:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sorry @SouravGhosh, I didn't quite get you.
  
  – Rai
  Mar 22 at 8:52
  

  
  
  
  

add a comment | 

1

"Why does pointer to an address of a double variable not work in C?"

No. It works.

double a=7; printf("%d",*&a); //It prints an address

• It's just printing the data at the address specified by &a. The * is for dereferencing the address.

And if you are saying that it's not working because it isn't printing the value of double a, it's simply because you are not using the right format specifier.

• 
  

  %d is the format specifier for int. If you want to print a double use %lf or %f.

  
  
  

  double a=7; printf("%lf",*&a); //This will print the double value

  
  

share|improve this answer

edited Mar 22 at 8:51

answered Mar 22 at 6:59

RaiRai

1,1424823

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  No. It works...that's an overstatement.
  
  – Sourav Ghosh
  Mar 22 at 7:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Sorry @SouravGhosh, I didn't quite get you.
  
  – Rai
  Mar 22 at 8:52
  

  
  
  
  

add a comment | 

"Why does pointer to an address of a double variable not work in C?"

No. It works.

double a=7; printf("%d",*&a); //It prints an address

• It's just printing the data at the address specified by &a. The * is for dereferencing the address.

And if you are saying that it's not working because it isn't printing the value of double a, it's simply because you are not using the right format specifier.

• 
  

  %d is the format specifier for int. If you want to print a double use %lf or %f.

  
  
  

  double a=7; printf("%lf",*&a); //This will print the double value

  
  

share|improve this answer

edited Mar 22 at 8:51

answered Mar 22 at 6:59

RaiRai

1,1424823

double a=7; printf("%d",*&a); //It prints an address

share|improve this answer

edited Mar 22 at 8:51

answered Mar 22 at 6:59

RaiRai

1,1424823

answered Mar 22 at 6:59

RaiRai

1,1424823

answered Mar 22 at 6:59

RaiRai

1,1424823