\Hi there)
Please tell me - how to ignore new line character in string during printing it by printf() -
char* str= "123\n4567";
printf("%s", str); // how to ignore here?
big thanks in advance)
Do you mean this:char* str= "123\\n4567";
printf("%s", str); // how to ignore here? to print 123\n4567
no - I want to escape \n (new line) character in string for ctime=
char* timeme()
{
const time_t timer = time(NULL);
return ctime(&timer);
}
I'm sorry... but I don't think we're understanding what you want to do...
If you have a string like 123\n456 and feed that to printf() or just about any other C display command, it's going to print as:
123
456
Now what do you want to see?
123456
If that's the case then you need to search for the \n which will probably exist in the string as either 0x0A and/or 0x0D and delete them before printing.
You could use a trim type function to remove white space from the time string.
#include <stdio.h>
#include <time.h>
char *trim(char *S)
{
while (*S == 32 || *S == 9 || *S == 10 || *S == 11 || *S == 13)
{
S++;
}
register int i = strlen(S);
while (i > 0 && (S[i - 1] == 32 || S[i - 1] == 9 || S[i - 1] == 10 || S[i - 1] == 11 || S[i - 1] == 13))
{
i--;
}
S[i] = '\0';
return S;
}
char *timeme(void)
{
const time_t timer = time(NULL);
return ctime(&timer);
}
int main(void)
{
char *s = timeme();
printf("The Date Time: %s trimed.\n", trim(s));
return 0;
}
char *timeme(void)
{
const time_t timer = time(NULL);
char *s = ctime(&timer);
*(s+24) = 0;
return s;
}
thanks for examples, guys)