Hello!
Can anyone explain line 28 for me? I was thinking *a would be replaced by *b, but it actually appends *a to *b. I know it is related to pointer address, but could not figure it out by myself. Thanks a lot!
Thanks!
Got the part for a[256], but not the stradd() function. I thought to use a = b, why shouldn't I?
I should have used *ptr1 and *ptr2 for the stradd() function to avoid confusion.
Thanks!
Got the part for a[256], but not the stradd() function. I thought to use a = b, why shouldn't I?
Consider it like this:
'char *' is like that variable 'a' up there. It doesn't tell you anything but where memory is.
The * operator dereferences it, (*a) in your code is like arraya[a] in my example.
The [] operator also dereferences it, like a[2] in your code would be like arraya[a+2] in my example.
Now, imagine that all memory in your process is all one great big array, so that 'a' and 'b' are simply different locations in the same big block. Altering those numbers doesn't change the memory they refer to.
Definitely I did not catch the basic work of pointer in C. I am falling into the same pitfall of C most people did, but I want get out as soon as possible. Needs more reading.
Thanks a lot!
This has helped me with pointers and pointers arithmetic: http://www.eskimo.com/~scs/cclass/notes/sx10.html but i can't say i have mastered it yet, au contraire. You probably know that there is a stdlib function called "strcat" ?
hi guys, I have this question. I am creating an script to that read a text file(.ini) with the list of the patterns to find for example:
EPMS_VO
EMPS_PARTS
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for i in `cat /home/linux1/xxxxxxx/f1.txt`;
do
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done
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it showing output
ravi
sankar
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Hello friends!
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Hi Friends,
I have a requirement
I need to concatenate the below two strings.
String 1 = /@jobid_at_
String 2 = value stored in ORACLE_SID
String 3 = string1 concatenated with String 2.
Please let me know how should i do it in UNIX.
Thanks, (2 Replies)