Comments for EmbLogic

Comment on Query: by sampath.kumar

sampath.kumar — Mon, 07 Oct 2013 04:00:03 +0000

At Preprocessing stage the expansion of inline function is done.

Comment on problem by jatinder.kumar

jatinder.kumar — Thu, 03 Oct 2013 17:56:35 +0000

*p get memory as read only. So when you printf(“%c\n”,–*p–); mesn you are goinh to change the contents of read only memory so that why you are getting segmantion fault

Comment on C: JATINDER KUMAR by jatinder.kumar

jatinder.kumar — Thu, 03 Oct 2013 17:52:05 +0000

How come?? mean on what basis your answer is?
Is there any change when complier optimize the both?? i even see the assmebly code of both and both are same . So on what basis i can say which one is faster

Comment on pointers by jasmeen

jasmeen — Thu, 03 Oct 2013 16:47:34 +0000

you can use instead of getting the input from stdin(scanf)

Comment on pointers by abhishek.jain

abhishek.jain — Thu, 03 Oct 2013 11:41:52 +0000

thnks mam ,
what is use of commamnd line arguments ?

Comment on Static linking of libraries. by jasmeen

jasmeen — Thu, 03 Oct 2013 09:10:15 +0000

No , loader has no role while linkiing

Comment on pointers by jasmeen

jasmeen — Thu, 03 Oct 2013 09:09:13 +0000

argc is number of command line arguments
and argv[0] points to name of file , argv[1] points to first command line argument and vice versa
for eg
u execute
./a.out 1 2 3 4
argc=4
argv[0]= ./a.out
argv[1]=1
argv[2]=2
argv[3]=3
argv[4]=4

Comment on problem by jasmeen.kaur

jasmeen.kaur — Mon, 30 Sep 2013 04:49:42 +0000

if it is — then –*p means u are changing the value at adress p(which is readonly) which will give you segmentation fault .
*p– means you are decrementing address i.e now it will point to ‘e’

Comment on problem by jasmeen.kaur

jasmeen.kaur — Sun, 29 Sep 2013 17:25:40 +0000

Can u pls tell what is written before *p in printf statement

Comment on Doubt: by taruna

taruna — Sun, 29 Sep 2013 08:01:41 +0000

The output will be:
This is
http://www.emblogic.com
website

and the first argument is not const char* but it is char*
Whenever we need to parse the same string , first argument in strtok() should be null.

Comment on Doubt: by msiddarth

msiddarth — Fri, 27 Sep 2013 18:12:10 +0000

strtok() will replace a the characters in the original string with a NULL each time the function is called using the same string, so the original string is modified by the use of strtok().
The second argument to strtok() can be changed at any time to a different delimiter.
Only single characters are considered to be delimiters.
The below example will illustrates the usage of strtok.
#include
#include

int main()
{
const char str[80] = “This is – http://www.emblogic.com – website”;
const char s[2] = “-”;
char *token;

/* get the first token */
token = strtok(str, s);

/* walk through other tokens */
while( token != NULL )
{
printf( ” %s\n”, token );

token = strtok(NULL, s);
}

return(0);
}

the output of the above program will be:
This is
http://www.tutorialspoint.com
website

Comment on problem by Ram D

Ram D — Wed, 11 Sep 2013 16:58:25 +0000

run following code in ur c99 compiler and compaire with the above one

#include
#include

void main()
{
char *p=”abcdef”;
for(int i=0;i<=6;i++)
{
p = p + (strlen(p)-i);
// printf("%c\n",–*p–);
// printf("%c\n",–*–p);
// printf("%c\n",–*(p–));
printf("%p\n",p);
printf("%c\n",*p);
}
}

Comment on C: JATINDER KUMAR by john McArthur

john McArthur — Sat, 31 Aug 2013 14:30:30 +0000

second one

Comment on c: JATINDER KUMAR by msiddarth

msiddarth — Tue, 27 Aug 2013 05:28:43 +0000

You can’t just cast an array to a structure because they’re simply incompatible types. Due to memory alignment constraints, the compiler needs to insert padding between the fields of a structure, so the members are not located at the memory addresses you may expect. Solutions:

Portable but slower/harder to do manually (preferred): copy manually the fields of the structure to the array.
Shorter to write but GCC-specific: use the __attribute__((packed)) keyword to make GCC not introduce padding between struct fields.

Comment on c: JATINDER KUMAR by ritesh

ritesh — Tue, 27 Aug 2013 01:12:45 +0000

short integer is of 2 byte.. we have stored the add of char array in structure which contain two member of unsigned short type.. so lt->a will pick the first two bytes of your character array which is ch={1,2,3,4,5}
so lt->a will points to 1st two byte which contains integer value 1,2 and bit representation of these two is
00000010 00000001 and converting into decimal will give you 513 and similarly lt->b will take next two integer 3,4 and bit representation is 00000100 00000011=1027