Using memset to initialize?

Using memset to initialize?

Bob Keane

Member #7,342

June 2006

Thinking in C++ Volume I Second Edition said:

The Standard C library function memset( ) (in <cstring>) is used for convenience in the program above. It sets all memory starting at a particular address (the first argument) to a particular value (the second argument) for n bytes past the starting address (n is the third argument). Of course, you could have simply used a loop to iterate through all the memory, but memset( ) is available, well-tested (so it’s less likely you’ll introduce an error), and probably more efficient than if you coded it by hand.

Is using memset the same as:

int array[5];
for(int i=0, i,5; i++)
array[i]=0;

My confusion is how does memset know where to stop?

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Goalie Ca

Member #2,579

July 2002

void * memset ( void * ptr, int val, size_t num );

ptr is starting address, num is number of bytes, val is what to set it to. Use sizeof.

in c++ you should be using std::fill instead.

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Timorg

Member #2,028

March 2002

void * memset ( void * ptr, int value, size_t num );

The 'num' parameter tells it how many bytes to set to to 'value'.

The equivalent code would be

int array[5];
memset(array, 0, sizeof(int) * 5);

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Bob Keane

Member #7,342

June 2006

Timorg said:

memset(array, 0, sizeof(int) * 5 )

That is what I was looking for. The book left the number of elements out of the example.

Thomas Fjellstrom

Member #476

June 2000

Or if its always a static array: memset(array, 0, sizeof(array));

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anonymous

Member #8025

November 2006

Actually the book's code is alright. Read it more carefully.

If you are declaring an array, you can just zero-initialize it right away:

int arr[5] = {};

When the array is a class member, you can zero-initialize it in the constructor:

class X
{
     int arr[5];
public:
     X(): arr() {}
};

In the book's example, it is used in another method, though, so some function is needed to zero out the memory. Why it chooses memset over std::fill or std::fill_n is somewhat unclear, though.

LennyLen

Member #5,313

December 2004

Bob Keane said:

The book left the number of elements out of the example.

Bob Keane said:

It sets all memory starting at a particular address (the first argument) to a particular value (the second argument) for n bytes past the starting address (n is the third argument).

Bob Keane

Member #7,342

June 2006

In my defense, it says n bytes past the first argument, not elements. That can be confusing to a newb.