12081 - Reduced ID Numbers

[ayon]

thanks Krzysztof Duleba, but i cannot understood the line

you just have to know what the second argument is of char type.

can you clarify in much easier way. why initializing an array with 0 is working, but not other values? can an array of structure be initialized directly with memset?

[misof]

thanks Krzysztof Duleba, but i cannot understood the line

you just have to know what the second argument is of char type.

can you clarify in much easier way. why initializing an array with 0 is working, but not other values? can an array of structure be initialized directly with memset?

the syntax is
memset( P, C, S );
where:
- P is a pointer into memory
- C is a variable of type char
- S is an integer variable (size)
memset takes the S bytes of memory starting at P and fills them with the char C.

As most of the variables use more than one byte (e.g., an int uses 4 bytes of memory), it is (in general) impossible to use memset for initialization of arrays.

There are some special cases where this is possible. For example, if all 4 bytes of an int are zero, the value of that int is zero. Thus if you use memset with C=0 on an array of ints, it will set all the ints to zero.

[ayon]

at last, i understand everything

Code: Select all

#include <stdio.h>
#include <string.h>

void main()
{
   int a;
   memset(&a, 1, sizeof(a));
   printf("%d\n", a);   
}

this code prints 16843009, because the 4 bytes are allocated like below:
00000001 00000001 00000001 00000001 = 16843009(decimal)

thanks to everyone

[Martin Macko]

There are some special cases where this is possible. For example, if all 4 bytes of an int are zero, the value of that int is zero. Thus if you use memset with C=0 on an array of ints, it will set all the ints to zero.

Very usefull is also the case if all 4 bytes of an int are -1 (11111111 binary), the value of that int is -1. Thus

Code: Select all

int P[100];
memset(a,-1,4*100);

fills P with -1s.