Converting char array to double in C?

Help with C++ Char array?

• 1. Write a program that initialises a char array {a, b, c}. Write a loop to print this array out on the screen without the NUL character. 2. Modify the program so that it prints the array out in reverse order (HINT: have one loop that searches for the terminator – then use the count found in a second loop to count backwards from there back to zero) 3. Write a program that reads two strings from the keyboard one after another. The program should then compare each location in the two arrays to see if they contain the same value. If all of the values are the same right through the string the program should print out “SAME” otherwise it should print out “DIFFERENT”. I have this code: int main() { char str[] = "abc"; int i = 0; { while(i < 3) { printf("%c\n", str[i]); i++; } system("pause>nul"); } } That covers question 1. For question 2 I would use: int i = 0; { while(i =0) printf("%c\n", str[i]); i--; But I think my tutor is asking something different here?

• Answer:

  include <stdio.h> int main() { char str[] = "abc"; int i = 0; while(i < 3){ printf("%c", str[i]); i++; } printf("\n"); system("pause>nul"); } That covers question 1. For question 2 I would use: int i = 0; while(str[i++]) ; i-=2; while(i) printf("%c", str[i--]); i--; part 2.............. #include <stdio.h> int main(void){ int i; char str[]="abc"; i=0; while(str[i++]) ;// the while loop stopped when we hit the terminating 0 of the string i--;// but we then added one to i so it is now two past the end of the string // so we decrement i to point to the terminating 0 while(i) printf("%c", str[--i]); // its --i so we start with the last valid character of the string // and stop when i is zero (after printing the zeroth character os str[] that is, str[0] printf("\n"); // print the new line return 0; } part three is left as an exercise for the student...