factorial program in C

[bonnieemmonsb]

I code in RPG but recently took a software testing test with that included the following code:

n! = n*((n-1)!)
int f=1;
if (n<0){
fprintf(stderr, 'factorial; Negative argument.'n';
}else if ((n==0) ||(n==1)){
f=1;
}else {
f=n*factorial(-n);
}
return(f)
}

what does the statement: {else if ((n==0)||(n--1)){ mean? The question is what inputs represent test cases that would give decision coverage with the minumum number of test cases? The answer is -1, 1, 5. I understand the -1 and the 5 but not the positive 1. I thought the answer would be -1, 0, 5. I code in RPG not in C so I attributed my incorrect answer to my lack of knowledge of C. Any help would be much appreciated.

[oler1s]

> what does the statement: {else if ((n==0)||(n--1)){ mean?
Else if (n is 0) or (n is 1):

The answer is -1, 1, 5. I understand the -1 and the 5 but not the positive 1. I thought the answer would be -1, 0, 5. I code in RPG not in C so I attributed my incorrect answer to my lack of knowledge of C.

I doubt your answer was wrong because of not knowing C. Actually, between their answer and your answer, I would favor yours. They have 1 and 0 as a base case for factorial, but it is possible to implement factorial with only 0 as a base case. For that reason, 0 should be tested, and 1 can be omitted.

[weaknessforcats]

It means that both 0! and 1! are 1.

Without this test, the recursion would be ininfite.

[svlsr2000]

by the way you should be having f=n*factorial(n-1); instead of f=n*factorial(-n);

[Ganon11]

Unless he forgot to type in an extra - making it f=n*factorial(--n);