Why is my C integer printing as a non-printable character?

[jd12345]

Ok so I was trying something in C when i found something i don't understand:-
code is:-
int c = 1;
printf("%c\n",c);
(not writing the include and return 0 stuff)

so when i run this it gives an output of a 2x2 box like this:-
0 0
0 1

whats happening. When i put the 1 in commas like this int c= '1' then it gives an ouput 1. Thats good but when i don't put the comma it gives this box, why?

 

[Mark44]

Ok so I was trying something in C when i found something i don't understand:-
code is:-
int c = 1;
printf("%c\n",c);
(not writing the include and return 0 stuff)

so when i run this it gives an output of a 2x2 box like this:-
0 0
0 1

whats happening. When i put the 1 in commas like this int c= '1' then it gives an ouput 1. Thats good but when i don't put the comma it gives this box, why?

I don't understand what you're talking about with the 2x2 box...

In C a character is a number: specifically, the value of a character is its ASCII code. Some characters, such as those whose ASCII codes are in the range 0 - 31, are nonprinting control characters.

There is a difference between the number 1 and the numeric character '1'. The value of the first is 1, of course, but the value of '1' is 49.

You can convince yourself of this by running this code.

printf("%c\n", 1);  // You won't see anything on the screen.
printf("%d\n", '1'); // Should print 49.

BTW, this-- , -- is a comma. This -- ' -- is a single quote.

 

[jd12345]

oh sorry about the single quote and comma error

printf("%c\n",1); prints the box.
I tried it - on linux (ubuntu) with GCC (if that matters)

 

[Adyssa]

The 'box' you are seeing is what happens when you print one of the non-printable ASCII chars.

int c = 1; // assign a new int the value 1
int c = '1'; // assign a new int the integer value of the character '1' (which is 49)

printf("%c", c); // print the character representation of the integer variable c
printf("%d", c); // print the integer representation of the integer variable c

Have a look at the ASCII table. Notice that the integer 1 represents the SOH (start of heading) character which is not printable. If you want to print the number one as stored in your integer variable, you should use the %d format specifier not %c.

 

[LudusRex]

There are a few potential reasons why your C integer is printing as a non-printable character. One possibility is that you are using the wrong format specifier in your printf statement. The %c format specifier is used for printing characters, not integers. Therefore, when you pass an integer to it, it will try to interpret it as a character and print the corresponding ASCII value. In your case, the ASCII value for the integer 1 is a non-printable character, which is why you are seeing the box instead of the number 1.

To print an integer, you should use the %d format specifier. This will convert the integer to its decimal representation and print it as a number. Alternatively, you can use the %i format specifier, which will print the integer in either decimal, octal, or hexadecimal format depending on the input value.

Another possible explanation for the box output could be a mismatch between the data types you are using. In your code, you have declared the variable c as an integer, but you are passing it to a character format specifier. This can lead to unexpected results as the printf function will try to interpret the data as per the format specifier, which may not match the actual data type of the variable.

In summary, to avoid getting non-printable characters when printing integers, make sure to use the correct format specifier and ensure that the data type of the variable matches the format specifier being used.