C programming passing a char array into a function - c

Question

I am using a function to parse through userID, and paswd and some error checking. The function is called from my main()... However when executed only the first 4 characters of my UserID and Pswd are successfully extracted. I am new to C-programming and coming from C# I am not sure where I am going wrong. This should be fairly easy, can someone point me in the right direction?
static void func1(int argc, char *argv[], char *UserID[30], char *Psw[30])
{
strncpy(UserID, argv[1], sizeof(UserID));
strncpy(Psw, argv[2], sizeof(Psw));
}
int main(int argc, char *argv[])
{
char UserID[30];
char Psw[30];
func1(argc, argv, UserID, Psw);
}
Also, just to point out, if I don't use the external function, and have all the code in my main func then it works.
EDIT:-
Figured out the issue:-
static void func1(int argc, char *argv[], char *UserID, char *Psw)
{
strncpy(UserID, argv[1], UserIDMaxSize);
strncpy(Psw, argv[2], PswMaxSize);
}
int main(int argc, char *argv[])
{
char UserID[UserIDMaxSize + 1]; /* max val defined in a header file */
char Psw[PswMaxSize + 1]; /* max val defined in a header file */
func1(argc, argv, UserID, Psw);
}
sizeof doesnt work quite as I expected it to.. it was reading the size of my pointer which is always 4 chars by default.

Answer 1

I guess your pointer has a size of 4 Byte. therefore you only read 4 chars.

Answer 2

Pass the array size to the function
static void func1(int argc, char *argv[], char *UserID, size_t UserIDSize,
char *Psw, size_t PswSize)
{
if (argc> 1) strncpy(UserID, argv[1], UserIDSize);
if (argc> 2) strncpy(Psw, argv[2], PswSize);
}
int main(int argc, char *argv[])
{
char UserID[30] = {0};
char Psw[30] = {0};
func1(argc, argv, UserID, sizeof UserID, Psw, sizeof Psw);
}
To insure the destination arrays are null character terminated, suggest strncat() --> "A terminating null character is always appended to the result." strncpy() has too many problems, it does not always result in an array with a null character.
static void func1(int argc, char *argv[], char *UserID, size_t UserIDSize,
char *Psw, size_t PswSize) {
UserId[0] = '\0';
// if (argc> 1) strncat(UserID, argv[1], UserIDSize);
if (argc> 1) strncat(UserID, argv[1], UserIDSize - 1);
Psw[0] = '\0';
// if (argc> 2) strncat(Psw, argv[2], PswSize);
if (argc> 2) strncat(Psw, argv[2], PswSize - 1);
}
[Edit]
Corrected code - off by 1

Answer 3

TL;DR
The sizeof isn't doing what you're expecting. Try using strlen instead.
You're only getting 4 characters copied because sizeof(char*[N]) for any N is just going to be the size of a pointer. On your platform, a pointer must be 4 bytes (32 bits).
I think you actually mean to pass the base-address of the array into the function, but in that case your types aren't quite right. Your compiler should be warning you about this. You should remove the * from your last 2 argument types:
static void func1(int argc, char *argv[], char UserID[30], char Psw[30])
That should get rid of the warning, and it should actually make sizeof behave correctly as well (since sizeof(char[30]) is 30). However, it's very easy to make mistakes with sizeof since the behavior with a char* and char[] are different... I'd prefer using strlen (or strnlen if you want to avoid possible buffer overflows) here instead, which will simply tell you how many non-null characters you have.
Using strnlen rather than sizeof would also help to tip you off that your parameter types are wrong, since it will complain that you're trying to pass a char** to a function that expects a char*.

Answer 4

Solution
#include <stdio.h>
#include <string.h>
void func1(int argc, char *argv[], char *UserID, char *Psw)
{
strncpy(UserID, argv[1], strlen(argv[1]));
strncpy(Psw, argv[2], strlen(argv[2]));
printf("DATA: %s \n",UserID);
printf("DATA1: %s \n",Psw);
}
int main(int argc, char *argv[])
{
char UserID[30];
char Psw[30];
printf("argv1 %ld \n",strlen(argv[1]));
printf("argv2 %ld \n",strlen(argv[2]));
func1(argc, argv, UserID, Psw);
}