make a char* which will hold newline

[swagatob]

I want to copy '\n' (i.e a newline) into a
char *src
(i.e to a pointer to a character). The apparent use will be to chcek this string with other strings I have for newlines. I need to keep it char* since my other strings are defined such.

char *newLineTest;
strcpy(newLineTest, '\n');

gives me a seg error saying that SIGSEGV,

any idea?

[Salem]

You need to do
src = malloc( some_number_of_bytes );

Plus strcpy takes strings, not characters

[kermit]

You have not allocated any space to store the newline character. If you must use char* then you need to use malloc. Otherwise, do something like:

char src[size_you need];

[swagatob]

hi
I used something llike this.. gives me SIGSEV, out of bound error

    char *newlinetest=malloc(sizeof(4));
    newlinetest='\n';

Apprently the strings that it will be comparing with can be as big as 100 chars. hence I also tried

    char *newlinetest=malloc(sizeof(100*sizeof(char)));
    strcpy(newlinetest,'\n');

both gave me same error

[Salem]

>strcpy(newlinetest,'\n');
Read!
'\n' isn't a string

strcpy(newlinetest,"\n");

Use double quotes

[Dave_Sinkula]

#include <stdio.h>

int main()
{
   printf("sizeof(100*sizeof(char)) = %d\n", (int)sizeof(100*sizeof(char)));
   return 0;
}

/* my output
sizeof(100*sizeof(char)) = 4
*/

[edit]If you mean 100, say 100.

char *newlinetest = malloc(100);

[swagatob]

yeep fixed thanks all again. you guys are wonderful

 char *newlinetest=malloc(100*sizeof(char));
 strcpy(newlinetest,'\n');

[Salem]

Still looks like single quotes to me

[swagatob]

oh yeah... it should be double quotes ""

you know I am so tired.... that I am making more mistakes with time now.. damn