Why is 'n' parameter of snprintf ignored?

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I have found that the n parameter of snprintf() seems to be ignored in my code.

char asdf[10];
Serial1.println(snprintf(asdf, 2, "hello"));

This prints 5 when I would expect it to print 2. What is happening?

string

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edited 30 mins ago

Greenonline

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asked 1 hour ago

Westin

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up vote
1
down vote

favorite

I have found that the n parameter of snprintf() seems to be ignored in my code.

char asdf[10];
Serial1.println(snprintf(asdf, 2, "hello"));

This prints 5 when I would expect it to print 2. What is happening?

string

share|improve this question

edited 30 mins ago

Greenonline

1,92241639

asked 1 hour ago

Westin

82

New contributor

Westin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I have found that the n parameter of snprintf() seems to be ignored in my code.

char asdf[10];
Serial1.println(snprintf(asdf, 2, "hello"));

This prints 5 when I would expect it to print 2. What is happening?

string

share|improve this question

edited 30 mins ago

Greenonline

1,92241639

asked 1 hour ago

Westin

82

New contributor

Westin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

I have found that the n parameter of snprintf() seems to be ignored in my code.

char asdf[10];
Serial1.println(snprintf(asdf, 2, "hello"));

This prints 5 when I would expect it to print 2. What is happening?

string

string

share|improve this question

edited 30 mins ago

Greenonline

1,92241639

asked 1 hour ago

Westin

82

New contributor

Westin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 30 mins ago

Greenonline

1,92241639

asked 1 hour ago

Westin

82

New contributor

Westin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

share|improve this question

edited 30 mins ago

Greenonline

1,92241639

edited 30 mins ago

Greenonline

1,92241639

edited 30 mins ago

Greenonline

1,92241639

1,92241639

asked 1 hour ago

Westin

82

New contributor

Westin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

Westin

82

asked 1 hour ago

Westin

82

82

New contributor

Westin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Westin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Westin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
3
down vote



accepted

snprintf() will not write more than <size> (snprintf's 2d argument) characters to your buffer, but it does count (and discard the extra) characters it would have written, had there been space enough, and that is the number it returns. Yeah, it can be confusing!

See this snprintf() reference.

share|improve this answer

answered 47 mins ago

JRobert

9,14811035

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It would be helpful because you could snprintf to a very small buffer, note the number returned, then malloc a buffer of the appropriate size, and do it again. That way you know how many bytes to allocate.
  – Nick Gammon♦
  2 mins ago
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

A test sketch for the Arduino Uno:

char buffer[10];

void setup() 
 Serial.begin(9600);
 int n = snprintf(buffer, 2, "hello");
 Serial.println(n);
 Serial.println(buffer);


void loop() 

As @JRobert wrote, the "would have" is the key. As far as I know only the snprintf and the vsnprintf return a "would have" number.

I think the reason is to be able to tell if the string was truncated. Suppose the 'size' parameter is 25 and the format string is very long, then the return value can be tested against 25. If the return value was 26 (the "would have" number of bytes), then the string was truncated.

This information was not possible to retrieve when the "would have" number was not available.

share|improve this answer

edited 33 mins ago

answered 39 mins ago

Jot

1,771416

add a comment | 

up vote
0
down vote

For completion, the man page for fprintf states:

The snprintf() function shall be equivalent to sprintf(), with the
addition of the n argument which states the size of the buffer
referred to by s. If n is zero, nothing shall be written and s may
be a null pointer. Otherwise, output bytes beyond the n‐1st shall be
discarded instead of being written to the array, and a null byte is
written at the end of the bytes actually written into the array.

and, more relevant:

Upon successful completion, the snprintf() function shall return the
number of bytes that would be written to s had n been sufficiently
large excluding the terminating null byte.

share|improve this answer

answered 22 mins ago

Greenonline

1,92241639

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

snprintf() will not write more than <size> (snprintf's 2d argument) characters to your buffer, but it does count (and discard the extra) characters it would have written, had there been space enough, and that is the number it returns. Yeah, it can be confusing!

See this snprintf() reference.

share|improve this answer

answered 47 mins ago

JRobert

9,14811035

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It would be helpful because you could snprintf to a very small buffer, note the number returned, then malloc a buffer of the appropriate size, and do it again. That way you know how many bytes to allocate.
  – Nick Gammon♦
  2 mins ago
  
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

snprintf() will not write more than <size> (snprintf's 2d argument) characters to your buffer, but it does count (and discard the extra) characters it would have written, had there been space enough, and that is the number it returns. Yeah, it can be confusing!

See this snprintf() reference.

share|improve this answer

answered 47 mins ago

JRobert

9,14811035

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It would be helpful because you could snprintf to a very small buffer, note the number returned, then malloc a buffer of the appropriate size, and do it again. That way you know how many bytes to allocate.
  – Nick Gammon♦
  2 mins ago
  
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

snprintf() will not write more than <size> (snprintf's 2d argument) characters to your buffer, but it does count (and discard the extra) characters it would have written, had there been space enough, and that is the number it returns. Yeah, it can be confusing!

See this snprintf() reference.

share|improve this answer

answered 47 mins ago

JRobert

9,14811035

snprintf() will not write more than <size> (snprintf's 2d argument) characters to your buffer, but it does count (and discard the extra) characters it would have written, had there been space enough, and that is the number it returns. Yeah, it can be confusing!

See this snprintf() reference.

share|improve this answer

answered 47 mins ago

JRobert

9,14811035

share|improve this answer

share|improve this answer

answered 47 mins ago

JRobert

9,14811035

answered 47 mins ago

JRobert

9,14811035

answered 47 mins ago

JRobert

9,14811035

9,14811035

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It would be helpful because you could snprintf to a very small buffer, note the number returned, then malloc a buffer of the appropriate size, and do it again. That way you know how many bytes to allocate.
  – Nick Gammon♦
  2 mins ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It would be helpful because you could snprintf to a very small buffer, note the number returned, then malloc a buffer of the appropriate size, and do it again. That way you know how many bytes to allocate.
  – Nick Gammon♦
  2 mins ago
  
  

  
  
  
  

1

1

It would be helpful because you could snprintf to a very small buffer, note the number returned, then malloc a buffer of the appropriate size, and do it again. That way you know how many bytes to allocate.
– Nick Gammon♦
2 mins ago

It would be helpful because you could snprintf to a very small buffer, note the number returned, then malloc a buffer of the appropriate size, and do it again. That way you know how many bytes to allocate.
– Nick Gammon♦
2 mins ago

add a comment | 

up vote
3
down vote

A test sketch for the Arduino Uno:

char buffer[10];

void setup() 
 Serial.begin(9600);
 int n = snprintf(buffer, 2, "hello");
 Serial.println(n);
 Serial.println(buffer);


void loop() 

As @JRobert wrote, the "would have" is the key. As far as I know only the snprintf and the vsnprintf return a "would have" number.

I think the reason is to be able to tell if the string was truncated. Suppose the 'size' parameter is 25 and the format string is very long, then the return value can be tested against 25. If the return value was 26 (the "would have" number of bytes), then the string was truncated.

This information was not possible to retrieve when the "would have" number was not available.

share|improve this answer

edited 33 mins ago

answered 39 mins ago

Jot

1,771416

add a comment | 

up vote
3
down vote

A test sketch for the Arduino Uno:

char buffer[10];

void setup() 
 Serial.begin(9600);
 int n = snprintf(buffer, 2, "hello");
 Serial.println(n);
 Serial.println(buffer);


void loop() 

As @JRobert wrote, the "would have" is the key. As far as I know only the snprintf and the vsnprintf return a "would have" number.

I think the reason is to be able to tell if the string was truncated. Suppose the 'size' parameter is 25 and the format string is very long, then the return value can be tested against 25. If the return value was 26 (the "would have" number of bytes), then the string was truncated.

This information was not possible to retrieve when the "would have" number was not available.

share|improve this answer

edited 33 mins ago

answered 39 mins ago

Jot

1,771416

add a comment | 

up vote
3
down vote

up vote
3
down vote

A test sketch for the Arduino Uno:

char buffer[10];

void setup() 
 Serial.begin(9600);
 int n = snprintf(buffer, 2, "hello");
 Serial.println(n);
 Serial.println(buffer);


void loop() 

As @JRobert wrote, the "would have" is the key. As far as I know only the snprintf and the vsnprintf return a "would have" number.

I think the reason is to be able to tell if the string was truncated. Suppose the 'size' parameter is 25 and the format string is very long, then the return value can be tested against 25. If the return value was 26 (the "would have" number of bytes), then the string was truncated.

This information was not possible to retrieve when the "would have" number was not available.

share|improve this answer

edited 33 mins ago

answered 39 mins ago

Jot

1,771416

A test sketch for the Arduino Uno:

char buffer[10];

void setup() 
 Serial.begin(9600);
 int n = snprintf(buffer, 2, "hello");
 Serial.println(n);
 Serial.println(buffer);


void loop() 

As @JRobert wrote, the "would have" is the key. As far as I know only the snprintf and the vsnprintf return a "would have" number.

I think the reason is to be able to tell if the string was truncated. Suppose the 'size' parameter is 25 and the format string is very long, then the return value can be tested against 25. If the return value was 26 (the "would have" number of bytes), then the string was truncated.

This information was not possible to retrieve when the "would have" number was not available.

share|improve this answer

edited 33 mins ago

answered 39 mins ago

Jot

1,771416

share|improve this answer

share|improve this answer

edited 33 mins ago

edited 33 mins ago

edited 33 mins ago

answered 39 mins ago

Jot

1,771416

answered 39 mins ago

Jot

1,771416

answered 39 mins ago

Jot

1,771416

1,771416

add a comment | 

add a comment | 

up vote
0
down vote

For completion, the man page for fprintf states:

The snprintf() function shall be equivalent to sprintf(), with the
addition of the n argument which states the size of the buffer
referred to by s. If n is zero, nothing shall be written and s may
be a null pointer. Otherwise, output bytes beyond the n‐1st shall be
discarded instead of being written to the array, and a null byte is
written at the end of the bytes actually written into the array.

and, more relevant:

Upon successful completion, the snprintf() function shall return the
number of bytes that would be written to s had n been sufficiently
large excluding the terminating null byte.

share|improve this answer

answered 22 mins ago

Greenonline

1,92241639

add a comment | 

up vote
0
down vote

For completion, the man page for fprintf states:

The snprintf() function shall be equivalent to sprintf(), with the
addition of the n argument which states the size of the buffer
referred to by s. If n is zero, nothing shall be written and s may
be a null pointer. Otherwise, output bytes beyond the n‐1st shall be
discarded instead of being written to the array, and a null byte is
written at the end of the bytes actually written into the array.

and, more relevant:

Upon successful completion, the snprintf() function shall return the
number of bytes that would be written to s had n been sufficiently
large excluding the terminating null byte.

share|improve this answer

answered 22 mins ago

Greenonline

1,92241639

add a comment | 

up vote
0
down vote

up vote
0
down vote

For completion, the man page for fprintf states:

The snprintf() function shall be equivalent to sprintf(), with the
addition of the n argument which states the size of the buffer
referred to by s. If n is zero, nothing shall be written and s may
be a null pointer. Otherwise, output bytes beyond the n‐1st shall be
discarded instead of being written to the array, and a null byte is
written at the end of the bytes actually written into the array.

and, more relevant:

Upon successful completion, the snprintf() function shall return the
number of bytes that would be written to s had n been sufficiently
large excluding the terminating null byte.

share|improve this answer

answered 22 mins ago

Greenonline

1,92241639

For completion, the man page for fprintf states:

The snprintf() function shall be equivalent to sprintf(), with the
addition of the n argument which states the size of the buffer
referred to by s. If n is zero, nothing shall be written and s may
be a null pointer. Otherwise, output bytes beyond the n‐1st shall be
discarded instead of being written to the array, and a null byte is
written at the end of the bytes actually written into the array.

and, more relevant:

Upon successful completion, the snprintf() function shall return the
number of bytes that would be written to s had n been sufficiently
large excluding the terminating null byte.

share|improve this answer

answered 22 mins ago

Greenonline

1,92241639

share|improve this answer

share|improve this answer

answered 22 mins ago

Greenonline

1,92241639

answered 22 mins ago

Greenonline

1,92241639

answered 22 mins ago

Greenonline

1,92241639

1,92241639

add a comment | 

add a comment | 

Westin is a new contributor. Be nice, and check out our Code of Conduct.

 

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