Detailed Print – Computer Sir Ki Class


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Solved Problem!#CPP#2697

Problem Statement - Array to array data

Write output of the following code and explain its working.


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int a2a(char *a1,char *a2,int rev=1)
A function definition with character array reference of two strings and the third variable rev to mark a condition with default value as 1.
int i=0; char ch; //The declarations and initialisations
int n=strlen(a1); //Finding length of first array
This loop increments address of a1 (first character array) one by one and reads the value in character ch until it hits the null character. This is basically a loop traversing method using array address.
if(rev) a2[n-i-1]=ch;
If rev variable is set the loop will fill a2 (the second array as passed by pointer location) from the last location determined by n-i-1 and successively moved left.

else a2[i]=ch; 
If variable is not set it would be simple copy operation.

i++; This moves the loop to next index of a1.
return i; //This will basically return the number of times a1 has been traversed, which is basically length of both a1 and a2.

int main() {
char a[20]=”hello”;
char b[20]=””;
int c=a2a(a,b);
Since a2a is called without third parameter, it will set the default value of rev as 1 to the reverse of a will be copied to b.

This should print olleh5   which is array b which is reversed of array a followed by returned array length.

This will set third parameter as 0 which means normal array copy will happen from a to b.

This should print hello5 as b will now be just a copy of a and the figure 5 will be its length.

return 0;


  • This is a tricky program which can perform both string copy and string reversal quite easily.
  • This program traverses the loop by moving the address pointer location rather than by array position movement as usually done. Address incrementing method is usually more efficient in practical implementations.

Common Errors

  • Student need to take additional care in putting brackets in address increment line in the while comparison condition in the loop.