random element from array in c

Question

How can I select a random element from a character array in c ?

For instance:

char *array[19];
array[0] = "Hi";
array[1] = "Hello";

etc

I am looking for something like array[rand], where rand is the random integer number between o and the array's length(in this case 20) like 1, 2, 3 , 19 etc.

• JBL: You're looking after rand()
• BLUEPIXY: maybe char* array[20];
• svk: So, to be clear, 20 * rand() is not correct. Use rand() % 20. However, if your array is 20 elements long you need to fix your variable declaration! You're only allocating 19 elements there.
• pjs: @svk No, don't use % because the results can be non-uniform. Multiplying a U(0,1) value times 20 and flooring will give an int from 0 to 19, i.e., 20 elements all set for zero-based array indices.
• Michael Anderson: What @pjs is refereing to is this issue from the linux man pages: "The versions of rand() and srand() in the Linux C Library use the same random number generator as random(3) and srandom(3), so the lower-order bits should be as random as the higher-order bits. However, on older rand() implementations, and on current implementations on different systems, the lower-order bits are much less random than the higher-order bits. Do not use this function in applications intended to be portable when good randomness is needed. (Use random(3) instead.)"

Answer 1

To start things off, since you have an array of strings, not of characters, you have to declare it as char* array[19];

Then, you can declare the following (always useful) macro

#define ARR_SIZE(arr) ( sizeof((arr)) / sizeof((arr[0])) )

Last, you can choose arr[rand() % ARR_SIZE(arr)] (while keeping in mind that performing % on rand() is not the proper way to do get a random number within a range.

• Jonathan Leffler: On most systems, you will get the same answer each time you run the program unless you seed the random number generator with a different value each time. Check out srand(), but be aware that getting a decent seed is actually rather hard. And the drand48() family of PRNG functions is quite useful too, though you still have the seeding problem.
• levengli: That is a correct comment, however I was referring to the issue of selecting a random number within a range of numbers in such a manner as to remain evenly distributed.

Answer 2

int n = rand()%20;
printf("%s\n", array[n]);

Answer 3

You can try array[rand() % ARRAY_LEN] but you are going to get a single char and not a char*

and when you are doing array[0] = "Hi"; it's not correct since you are assigning to a single char a char*

or turn your char array[20] into a char *array[20] and you can assign a string of characters

• Rrjrjtlokrthjji: You are correct. I was getting seg fault before I modify the array into a pointer. (Upvote, thanks)

Answer 4

What you propose is the best solution there is - choose a random index and then use the element at this index. If your question is how to get a random integer, use the built-in function rand().

Answer 5

This can be done using rand in the c library stdlib.h

You can get a random number like this:

char random_elem = array[rand()%20];

and you can print it out like this:

printf("%d",array[rand()%20]);

• Peter Wolf: Thank you for your contribution, but this question already has few good-quality answers and your answer adds nothing new. How about focusing on some unanswered C questions?