What is the output of the following program? void main() { printf("1"); goto…

In C, the statement goto label; performs an unconditional jump to a point marked label: within the same function. A label must be followed by a colon, not a semicolon — label: marks a jump target, while label; would instead be read as an ordinary (and here, undeclared) expression statement. The instant execution reaches a goto, control moves straight to its target label; every statement written between the goto and the label is skipped, and normal top-to-bottom execution resumes from the label onward.

1. Control enters main and executes the first statement, printf("1"), so 1 is printed.

2. The next statement is goto xyz;, an unconditional jump to the label xyz:. Control transfers there immediately, so the intervening printf("2") never runs.

3. Execution resumes at xyz: and continues to the next statement, printf("3"), so 3 is printed.

4. No further statements remain, so the characters actually printed, in the order they were printed, form the complete output.

• If the goto line were removed, execution would fall straight through every statement in source order, printing all three characters — confirming that it is specifically the jump, not some other effect, that removes the middle print from the output.

• The label xyz: itself performs no printing; it only marks where execution resumes after the jump, so it contributes nothing extra to the output.

The two prints that actually execute, in order, produce the output 13.