Display highest value in array in C - c

Question

When I run this, ageProgMax comes out as 29 instead of what I want, 60. I did this for the smallest number of cafes that analysts drink and it worked but somehow this one doesn't.
int main()
{
char poste[] ={'P', 'P', 'O', 'A', 'P', 'A', 'P', 'P'};
int nbCafe[] ={5, 1, 3, 0, 5, 1, 0, 2};
int age[] ={25, 19, 27, 22, 49, 24, 60, 29};
int nbPers = sizeof(age) / sizeof(int);
int i;
int ageProgMax = 0;
for (i = 0; i < nbPers; i++)
if (poste[i] =='P' || age[i] > ageProgMax)
{
ageProgMax = age[i];
}
printf ("Max age of programmers : %d\n", ageProgMax);
return 0;
}
Any help?
Thanks

Answer 1

It's because of the || in your condition. Look at the condition you have set i.e. if (poste[i] =='P' || (age[i] > ageProgMax)). It says to store new value to ageProgMax if either of (age[i] > ageProgMax) or poste[i] =='P' becomes true. So for the last entry i.e. 29 even though (age[i] > ageProgMax) is false, the poste[i] =='P' is true and causes the overwrite of the ageProgMax's 60 with that of 29.
You can correct your program like this.
int main()
{
char poste[] ={'P', 'P', 'O', 'A', 'P', 'A', 'P', 'P'};
int nbCafe[] ={5, 1, 3, 0, 5, 1, 0, 2};
int age[] ={25, 19, 27, 22, 49, 24, 60, 29};
int nbPers = sizeof(age) / sizeof(int);
int i;
int ageProgMax = 0;
for (i = 0; i < nbPers; i++)
{
if (poste[i] =='P' && (age[i] > ageProgMax))
{
ageProgMax = age[i];
}
}
printf ("Max age of programmers : %d\n", ageProgMax);
return 0;
}

Answer 2

#define MAX(a,b) ((*(a)) > (*(b)) ? (a) : (b))
int *largest(int *arr, size_t size)
{
int *result = NULL;
if(arr && size)
{
if(size == 1)
{
retult = *arr;
}
else
{
result = MAX(arr, arr + 1)
arr += 2;
size -= 2;
while(size--)
{
result = MAX(result, arr);
arr++;
}
}
}
return result;
}
Returns NULL on error or pointer to the largest int.