MEMSET Doubt ?

Question

Hi,

Following piece of code works perfectly :-

#include <iostream>
#include <cstdio>

using namespace std;

int main(){
	int x[10];	
	memset(x, 0, sizeof(x));	
	return 0;
}

In the debugger, I can see that each element of array 'x' is 0 indeed. However when I do this:
memset(x, 34, sizeof(x));

Then each value of x is some random value, sometimes 10 digits long. Actually all I want to do is, fill the array 'x' with a particular value WITHOUT running for loop from 1 to 10 and manually substituting the value like this:

x[i] = 34;

I looked on the internet and found memset but it only works for 0.

Thanks

Amare.

Answer 1

It is not a random number. memset(x, 34, sizeof(x)); sets every byte in x to 34. Each int will be 0x22222222 = decimal 572662306, which should be the number the debugger showed you.

Why not use

for (int n=0; n<(sizeof(x)/sizeof(x[0])); n++) x[n] = 34?

Answer 2

Memset fills a byte array, but you want to fill an array consisting of int. Each integer consists of several bytes and each byte is set to 34. As a result the whole integer does not contain the expected value.

 

Try fill and fill_n functions declared in <algorithm>:

 

    fill_n(x, 10, 34);

 

Answer 3

Thanks hgn for the help.

Answer 4

Thanks Viorel_ ! Your suggested fill_n function is what I have been looking for.
Thanks for the help.