char *tempMonth;
char month[4];
month[0]='j';
month[1]='a';
month[2]='n';
month[3]='\0';
how to assign month to tempMonth? thanks
and how to print it out finally?
thanks
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6 Answers
In C, month == &month[0] (in most cases) and these equals a char * or character pointer.
So you can do:
tempMonth=month;
This will point the unassigned pointer tempMonth to point to the literal bytes allocated in the other 5 lines of your post.
To make a string literal, it is also simpler to do this:
char month[]="jan";
Alternatively (though you're not allowed to modify the characters in this one):
char *month="jan";
The compiler will automatically allocate the length of the literal on the right side of the month[] with a proper NULL terminated C string and month will point to the literal.
To print it:
printf("That string by golly is: %s\n", tempMonth);
You may wish to review C strings and C string literals.
1 Comment
paxdiablo
just some suggestions: change
= to == in the first line and make it clear that month == &month[0] mostly. Not so when using sizeof or certain other language features.If you just want a copy of the pointer, you can use:
tempmonth = month;
but that means both point to the same underlying data - change one and it affects both.
If you want independent strings, there's a good chance your system will have strdup, in which case you can use:
tempmonth = strdup (month);
// Check that tempmonth != NULL.
If your implementation doesn't have strdup, get one:
char *strdup (const char *s) {
char *d = malloc (strlen (s) + 1); // Allocate memory
if (d != NULL) strcpy (d,s); // Copy string if okay
return d; // Return new memory
}
For printing out strings in a formatted fashion, look at the printf family although, for a simple string like this going to standard output, puts may be good enough (and likely more efficient).
Comments
tempMonth = month
When you assign a value to a pointer - it's a pointer, not a string. By assigning as above, you won't miraculously have two copies of the same string, you'll have two pointers (month and tempMonth) pointing to the same string.
If what you want is a copy - you need to allocate memory (using malloc) and then actually copy the values (using strcpy if it's a null-terminated string, memcpy or a loop otherwise).
Comments
#include "string.h" // or #include <cstring> if you're using C++
char *tempMonth;
tempMonth = malloc(strlen(month) + 1);
strcpy(tempMonth, month);
printf("%s", tempMonth);
2 Comments
paxdiablo
Some would say that, if you use
printf or malloc, then you couldn't possibly be using C++ - you're stuck in purgatory awaiting full conversion :-)
Andrew Rasmussen
true. you can still include C libs though. you never know ;)
tempmonth = malloc (strlen (month) + 1); // allocate space
strcpy (tempMonth, month); //copy array of chars
Remember to:
include <string.h>
3 Comments
Andrew Cooper
There's a buffer overflow waiting to happen. Always use
strncpy, and make sure tempMonth points to properly allocated memory.paxdiablo
I didn't downvote you but I think your allocation should be for
strlen(month)+1. sizeof(tempmonth) is the size of the pointer.paxdiablo
Fixed and upvoted. You know, you don't have to leave your answers as they are if you want to change them (I think you can edit your own posts at any rep level).
You could do something like:
char *months[] = {"Jan", "Feb", "Mar", "Apr","May", "Jun", "Jul", "Aug","Sep","Oct", "Nov", "Dec"};
and get access
printf("%s\n", months[0]);
