Printing char by integer qualifier

Printing char by integer qualifier

By : Kagiri
Date : November 25 2020, 07:22 PM
I hope this helps . A characters that begins with \0 represents Octal number, is the base-8 number system, and uses the digits 0 to 7. So \08 is invalid representation of octal number because 8 ∉ [0, 7], hence you're getting implementation-defined behavior.
Probably your compiler recognize a Multibyte Character '\08' as '\0' one character and '8' as another and interprets as '\08' as '\0' + '8' which makes it '8'. After looking at the ASCII table, you'll note that the decimal value of '8' is 56.
code :

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Printing the memory representation of an integer using char* and unsigned char*

Printing the memory representation of an integer using char* and unsigned char*

By : Bertin Cano
Date : March 29 2020, 07:55 AM
wish help you to fix your issue The type is getting promoted to signed int when passed to the printf function. When a signed type is promoted, it's "sign-extended".
To an 8-bit signed char, the value 250 is equivalent to -6 (2's complement) and since it's signed, -6 is considered the "true" value. When this is extended to a signed integer (4 bytes), as is happening here, the value -6 is preserved (through the sign-extension), rather than the value 250. But -6 on a 4 byte value is fffffffa, whereas it's merely fa on a single byte.
How to construct a char array with absolute integer value(s) and equivalent char integer value(s) in C?

How to construct a char array with absolute integer value(s) and equivalent char integer value(s) in C?

By : Sergey Shlokov
Date : March 29 2020, 07:55 AM
this will help For example, I need a char array with contains the NUL '\0' character followed by char '1'. Initializing the char array as: , From the C++ Standard:
code :
char *str = "foo\0001bar";
char *str = "foo\0" "1bar";
From integer to char array of digits and printing

From integer to char array of digits and printing

By : Pierrot
Date : March 29 2020, 07:55 AM
Hope this helps You are type casting Integer to Character, So it will consider your int as a ASCII value of some character. And It will add that character to your array instead of your number. So try following:
code :
int baseChange (int base, int newBase, int num)
    int baseTen = toBaseTen(base, num); //give me a number on base 10
    char finalBase[24];
    int i;
    for (i = 0; baseTen != 0; i++)
        int remain = baseTen % newBase;
        finalBase[i] = (char)(remain+48); //insert the last digit to the array
        baseTen /= newBase;
    // Print finalBase from end to beginning in order to get the final number in the correct order:
    for (i--; i >= 0; i--)
        printf("i is: %d... %c", i, finalBase[i]);
    return SUCCESS_EXIT;
Printing char as integer

Printing char as integer

By : Coflo Cody Ferreira
Date : March 29 2020, 07:55 AM
Any of those help Largely for historic reasons, characters are small integers, and all integer arithmetic is done in the int type or wider. Software sometimes had to work with characters as integer codes and sometimes as things to be printed, so C developed with the char type being flexible.
Wherever an int or unsigned int may be used in an expression, including function arguments, you may use a char instead.
printing sum of integer and a char*

printing sum of integer and a char*

By : Beehive Web Solution
Date : March 29 2020, 07:55 AM
around this issue printf expects a pointer to char. "+2" means shift pointer beyond 2 positions.
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