CODEWITHSUNDEEP
carrayscharprintf
flexzican
flexzican

Reputation: 97

char array prints wrong value in C

I am stuck on the same code for quit some time now. I am trying to fill a char array with characters i read from a text file (ascii). But for some reason when i printf the char array it only displays the letter H.

Code:

void append(char c)
{
  //int len = strlen(cStr);
  cStr[iCounter] = c;
  cStr[iCounter + 1] = '\0';
  printf("char c:%c    char array%c\n",c,cStr);
}

The char array (cStr) is declared outside this function because i need to acces it from different functions. So is iCounter which is incremented every time it executes this function.

Any help would be appreciated.

Upvotes: 2

Views: 2189

Answers (2)

iabdalkader
iabdalkader

Reputation: 17332

You print one character with %c use string specifier %s instead:

printf("char c:%c    char array%s\n",c,cStr);

Note: iCounter is not actually incremented:

  cStr[iCounter++] = c;
  cStr[iCounter] = '\0';

Upvotes: 3

tomahh
tomahh

Reputation: 13661

Use %s to print string. %c is use to print unique character.

From printf man page

c

If no l modifier is present, the int argument is converted to an unsigned char, and the resulting character is written. If an l modifier is present, the wint_t (wide character) argument is converted to a multibyte sequence by a call to the wcrtomb(3) function, with a conversion state starting in the initial state, and the resulting multibyte string is written.

Upvotes: 1

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