Answer:
456 will be printed
Explanation:
The following shows a well formatted equivalent of the code snippet:
String str = "abc456"; //Line 1
int m = 3; //Line 2
while (m <= str.length() - 1) { //Line 3
if (Character.isDigit(str.charAt(m))) //Line 4
System.out.print(Character.toUpperCase(str.charAt(m))); //Line 5
m++; //Line 6
} //Line 7
Line 1 shows the declaration and initialization of a String variable str.
Line 2 shows the declaration and initialization of an integer variable m.
Lines 3 through 7 show a while loop.
The loop starts at m = 3, increments m by one at every cycle, and ends at m = 5 which is one less than the length of the String str. i.e 6 - 1 = 5
At each cycle, the loop checks if the character at index m of the str is a digit. If it is a digit, it is converted to an uppercase letter(which is irrelevant) and then printed to the console.
At m=3, the character is a digit which is 4
At m=4, the character is a digit which is 5
At m=5, the character is a digit which is 6
Therefore, 456 is printed to the console.
Note: I have assumed that the statement in the while loop of the code you wrote was a typo and as such I have changed it to the correct syntax. That is, I have changed ;
while(m =< str.length( ) - 1) to
while(m <= str.length( ) - 1) for the code to perform the intended function.
Otherwise, if it is the case that it wasn't a typo, then the code will neither compile nor run as it contains a syntax error.
Hope this helps!
