C programming2

Question

What will be output

int main()
{
  char str[10]="GATE 2016";
  int length=strlen(str);
  str[length]='\0';
  for(i=0;str[i];i++)
    printf("%c",str[i]);
  return 0;
}

Answer 1

Strlen returns the size of the string = 9 .

In the loop, it is printing one be one all the characters having the condition str[i] TRUE.

In C/C++, value 0 is false and any other value is TRUE. Also, Null character '\0' is guaranteed to have the value '0' (in the byte for '\0', 00000000 will be stored) in any character encoding that is used in the system as per C standard. So, the loop terminates at str[i] == ' \0 ' .

Output will be GATE 2016 . 

PS: Statement 9 - str[length] = '\0' is unnecessary as in C all string literals are automatically given '\0' at end by the compiler. So, "GATE 2016" will be stored in memory as 'G','A','T','E',' ','2','0','1','6','\0'.

Comments

Yes, but you do not need a strcpy there - string can be directly given to sscanf.

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sscanf reads the data from s rather than stdin.So, whenever it gets 0 it terminates the execution. rt?

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it works exactly as for scanf from standard input.

s

Matches a sequence of non-white-space characters; the next pointer must be a pointer to character array that is long enough to hold the input sequence and the terminating null byte ('\0'), which is added automatically. The input string stops at white space or at the maximum field width, whichever occurs first.

http://linux.die.net/man/3/scanf