Store an integer in char array in C

Question

Running the code below prints b = and i = 15.

int i = 15;
char b = (char) i;
printf("b = %c and i = %d\n", b, i);

How can I store this integer in the character? At the end I'm trying to have a char array of size 1024 which has i (15) as the first character and rest 0.

update: I tried :

int i = 15;
char buffer[1024];
snprintf(buffer, 10, "%d", i);
printf("buffer[0] = %c, buffer[1] = %c\n", buffer[0], buffer[1]);

And the result printed was:

buffer[0] = 1 , buffer[1] = 5

Solution 1

You did store the integer in the character, it's just that %c converts a character to its ASCII value. All ASCII values below 31 are non-printable.

If you run

printf("b = %d and i = %d\n", (int)b, i);

it will print 15.

If you want a representation of i as a string:

char buf[12]; //Maximum number of digits in i, plus one for the terminating null
snprintf(buf, 12, "%d", i);

This will store a string representation of i in buf.

Solution 2

Clarification:

The range of char is -128..127. The range of unsigned char is 0..255.

If capturing ASCII values is the goal, declaring buffer variable to unsigned char type seems to be more appropriate.

Solution 3

The problem here is, variable b already has a value 15 but since this does not constitute to a printable ASCII, using %c format specifier, you won't be able to see any output.

To print the value, use %hhd format specifier.

At the end I'm trying to have a char array of size 1024 which has i (15) as the first character and rest 0.

Well, you can define an array and assign values accordingly. Something like

#define SIZE 1024

char arr [SIZE] = {0}; //initialization, fill all with 0
arr[0] = 15;           //first value is 15

should do the job.

Solution 4

A char is an 8-bit unsigned value (0 - 255) and it does indeed store 15 in it, the problem is, that in ASCII table 15 means "shift in" non-printable character, and %c interprets the value as an ascii character.

char b = (char) i;
printf("b = %d and i = %d\n", b, i);

to get

b = 15 and i = 15

if you used i = 90 in your current code, this would be printed:

b = Z and i = 90

View more solutions

Comments

[Dmitri]

char isn't always unsigned.

[Nicky Mirfallah]

I tried: char buffer[1024]; snprintf(buffer, 1, "%d", i); printf("buffer[0] = %c, buffer[1] = %c\n", buffer[0], buffer[1]); But i'm still not getting 1 and 5 for buffer (i=15)

[Dmitri]

Shouldn't that be snprintf(buf, 10, "%d", i);?

[anonymoose]

@Dmitri Oops. Thanks for the catch.

[anonymoose]

@NickyMirfallah As Dimitri mentioned, I had a minor typo. char buffer[1024]; snprintf(buffer, 1024, "%d", i); printf("buffer[0] = %c, buffer[1] = %c\n", buffer[0], buffer[1]); should work. Note thatprintf("%s\n", buffer); will print out the whole string ("15" in this case).

[chux - Reinstate Monica]

ASCII values are in the range 0 to 127 and can be saved in char , unsigned chart, or signed char.

[chux - Reinstate Monica]

%hhd converts the integer argument to signed char before printing. char might not be signed char. IAC, printf("b = %dn", b); is sufficient unless rare char is same width/signness as unsigned.

[Nguai al]

Yes. But any ASCII value that is greater than 127 will require unsigned char type.

[chux - Reinstate Monica]

char b = (char) i; is fine. "You can't store Integer datatype in Character datatype" is not supported by the C spec. char is an integer type.

[chux - Reinstate Monica]

Hmmm, typical int can store the value of -2,147,483,648. Magic number 10 is insufficient for a string that may need size 12.

[chux - Reinstate Monica]

Since ASCII is defined in the range 0 to 127, "ASCII value that is greater than 127" is a false premise. Perhaps you are thing of various extensions to ASCII?

[Nguai al]

Yes. I am considering ASCII extensions.

[anonymoose]

@chux The comment next to the declaration was meant as an instruction to change it based on the range of values needed for the application, not a claim that 10 is the number needed. But good point. I'll update it.