Study the function:void t(char* S) { for (int i=0; i<strlen(S); i++) S[i] = S[i]+1;}This function is called as:char S[10]= “ABCD”;printf(“%s”, t(S));What is the output?a.Abcdb.Compile-time error.c.ABCDd.BCDE
Question
Study the function:void t(char* S) { for (int i=0; i<strlen(S); i++) S[i] = S[i]+1;}This function is called as:char S[10]= “ABCD”;printf(“%s”, t(S));What is the output?a.Abcdb.Compile-time error.c.ABCDd.BCDE
Solution
The function t takes a string S as input and increments each character in the string by 1. So, 'A' becomes 'B', 'B' becomes 'C', and so on.
However, there is a mistake in the question. The function t is of type void, which means it does not return anything. But in the printf statement, it is used as if it returns a string. This will cause a compile-time error.
So, the correct answer is b. Compile-time error.
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