Exit status -1
This is my code and I am not getting why is it showing exit status -1.
scanf("%s",c); //to take in the 1st string
scanf("%c",&d);//after string the newline char needs to get in the buffer
scanf("%s",e); //to take in another string
printf("%s\n%s",c,e); //to print both the strings
%s is only for strings
you've got 3 chars
and use scanf like so:
ampersand(&) is only for non-arrays like char, int, float
but not char, int, float etc...
@Yuvikiit I see, repl formatted your asterisks out.
Try one backwards tick mark ` for inline code, or three for code with multiple lines.
As for your code, when you declare a char pointer:
You need space to store stuff in it.
Arrays do this automatically because you tell it from the start how much space it should have like so:
char c; // space for 20 chars already made at compile
To do this with pointers:
OR you can steal the space made by an array if you don't want to use malloc
@mwilki7 Sankyu(Thanks) Sir for the reply, just let me know one more thing
Like why this works for a single string.
and how does it work:
I thought in the manner that being a character pointer
the pointer would store 1 character as I am entering and then will move to next byte for latter entry and as soon as I press enter it puts a '\0' at the end and completes the string,
And I also don't know like why this should be a problem even
is it because that the pointer has some garbage address stored in it and it is not a good programming practice to store our string at some garbage place though, the memory chunks might be continuous (as the char pointer
does that bit by default.)
let me know is it some compiler dependant thing because I tried the same thing in Codeblocks and it didn't work.
Thanks once again for the help.