My Loop Won't Check All Elements Of Array in C - c

Question

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void displayString (const char *sPtr);
void getString (char *[]);
int determinIfConvert (char);
int main ()
{
char originalString[11] = { 0 };
char convertedString[11];
getString (originalString);
displayString (originalString);
// this loop runs through the "originalString" to check for the char: 'a'
for (int i = 0; i < 11; i++) {
determinIfConvert (originalString[i]);
}
system ("pause");
}
void getString (char *a[]) // this function gets a string
{
printf ("enter 11 char string: \n");
scanf ("%s", a);
}
// this program displays the inputstring
void displayString (const char *sPtr)
{
for (; (*sPtr != '\0'); ++sPtr) {
printf ("%c", *sPtr);
}
}
int determinIfConvert (char *a)
{
if (a == 97) // this is a test condition. The goal is to
// check for all lowercase, but now i'm
// only entering "aaaaa"
{
printf ("Works"); // if it prints multiple"works"
// then i can continue my program
// but it only prints ONE "works" and freezes.
}
}
At the moment I have a problem with my For Loop in main() not finishing. The goal is to enter a string of characters, and then check for lowercase ones. This will be done with the function DeterminIfConvert(char). However, when I run through the loop element by element, it freezes after the second element. My test data is "aaaa" and it prints the "aaaa," so I know that my first two functions work just fine. I get to the loop, it goes through the first element, prints "works" and then freezes. :/

Answer 1

Multiple mistakes
void getString(char *a[])
should be
void getString(char a[])
Since you're sending the base address of an array of char, not an array of pointer to char
char *a[]; // array of pointer to char
char a[]; // array of char
int determinIfConvert(char *a)
should be
int determinIfConvert(char a)
Since you're sending a char, not a pointer to char
char * a; // pointer to char
char a; // char
NOTE:
Use the standard definition of main()
int main(void) //if no command line arguments.

Answer 2

If you are inputting an 11-char string, then you should be doing:
char originalString[12] = { 0 };
This is because you need 1 more character to store the null character '\0'.
That is probably why in your function getString(...), the pointer exceeds the array bounds and might invoke undefined behavior.
Finally, your function prototype for getString(...) should be
void getString(char a[]); //without the *

Answer 3

In addition to the other answers, you have several other areas where you can improve your code.
Avoid using magic numbers in your code (e.g. 11). Instead define a constant for the maximum characters in your string #define MAXC 11 or you can use an enum instead enum { MAXC = 11 };
As it currently sits, you do not protect against overflowing your 11 character array (which means your user can enter no more than 10 characters plus room for the nul-terminating character). To protect against the user entering something more than 10, you should use a field-width specifier with scanf:
scanf ("%10s", a);
That doesn't solve your problems with scanf. You must check the return every time to insure the expected number of conversions takes place, e.g.:
if (scanf ("%10s", a) != 1) {
fprintf (stderr, " -> error: invalid input.\n");
exit (EXIT_FAILURE);
}
That's better, but using %s, you cannot read a string containing whitespace, and you are still leaving a trailing '\n' in the input buffer. If the users enters "my dog", you store "my" only. To fix part of the problem you can use a format specifier of "%10[^\n]%*c". However, you must protect against an endless-loop if the user presses [Enter] without other input. To resolve all issues, and prevent leaving the trailing newline in the input buffer, you can use something like:
int getString (char *a) // this function gets a string
{
int c, rtn = 0;
printf ("enter string (10 char or less): ");
while ((rtn = scanf ("%10[^\n]%*c", a)) != 1) {
if (rtn == EOF)
break;
fprintf (stderr, " -> error: invalid input, try again..\n");
printf ("enter string (10 char or less): ");
/* flush input buffer - to avoid endless loop */
while ((c = getchar()) != '\n' && c != EOF) {}
}
return rtn;
}
All of which expose the difficulties using scanf for user input. A better approach may be to use fgets (or getline) to read the complete line of input.
Regardless whether you use scanf or fgets, etc.. you must take a bit of time and care in writing your input handlers to insure you try and cover all ways a user could muck up input. Below fgets is used just to present an alternative. You should also choose a return type that allows you to tell whether you have successfully received input or not. It might as well be a useful return such as the length of the input taken, etc..
The remainder of your level of pointer indirection issues have been addressed by other answers. Putting it all together, you could do something like:
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAXC 11
void displayString (const char *sPtr);
int getString (char *);
int determinIfConvert (char);
int main (void)
{
char originalString [MAXC] = "";
// char convertedString[MAXC] = ""; /* currently unused */
if (!getString (originalString)) {
fprintf (stderr, "error: getString failed.\n");
return 1;
}
displayString (originalString);
// this loop runs through the "originalString" to check for the char: 'a'
for (int i = 0; i < 11; i++) {
determinIfConvert (originalString[i]);
}
system ("pause");
return 0; /* main() is type 'int' and returns a value */
}
int getString (char *a) // this function gets a string
{
char *p = a;
int c;
size_t len = 0;
printf ("enter string (10 char or less): ");
for (;;) {
p = fgets (a, MAXC, stdin);
if (!p) break; /* handle [CTRL+D] */
if (*p == '\n') { /* handle empty str */
fprintf (stderr, " -> error: invalid input, try again..\n");
printf ("enter string (10 char or less): ");
continue;
}
/* trim newline/flush input buffer */
len = strlen (p);
if (len && a[len - 1] == '\n')
a[--len] = 0;
else /* user entered more than 10 chars */
while ((c = getchar()) != '\n' && c != EOF) {}
break;
}
return (int) len;
}
// this program displays the inputstring
void displayString (const char *sPtr)
{
for (; *sPtr; sPtr++) {
printf ("%c", *sPtr);
}
putchar ('\n');
}
int determinIfConvert (char a)
{
if (a == 97)
printf ("Works\n");
return 0;
}
Example Use/Output
$ ./bin/getdispstr
enter string (10 char or less): my dog has fleas
my dog has
Works
$ ./bin/getdispstr
enter string (10 char or less):
-> error: invalid input, try again..
enter string (10 char or less): my dog has fleas, my cat has none.
my dog has
Works
With CTRL+D (EOF)
$ ./bin/getdispstr
enter string (10 char or less): error: getString failed.
There are many ways to do this, this is just an example. Look over all the answers and let me know if you have questions.

Answer 4

This
char originalString[11] = { 0 };
followed by this
for (int i = 0; i < 11; i++)
{
determinIfConvert(originalString[i]);
}
is causing the problem. You see the array of char does not have elements post index 0. And yeah I believe what you are trying to attempt with
getString(originalString); seems like you want to get originalString from user input which is not correctly executed in your case.

Answer 5

You pass object of type char to a function accepting char*
char originalString[11] = { 0 };
determinIfConvert(originalString[i]);
int determinIfConvert(char *a)

Answer 6

A string is nothing but a null terminated set of characters, so if you wish to have 11 characters in you string, you should be allocating 12 bytes to your
array, ie you may change :
char originalString[11] = { 0 };
to
char originalString[12] = "";
/* Here is the string is empty but because you use double quotes
* compiler understands that you are initializing a string, so '\0' is auto
* appended to the end of it by the compiler to mark the end of the string.
*/
So is the case with convertedString[11] change it to
char convertedString[12] = "";
Change
void getString(char *a[]);
to
void getString(char a[]); //char *a is also fine
Change
int determinIfConvert(char *a)
to
int determinIfConvert(char a) // You wish to check a character
You may wish to replace
scanf("%s", a);
with
fgets(a,12,stdin);
because scanf can't check for overflows but fgets can. Here you can have up to 11 characters in the string. If an overflow occurs, the rest of the input is trimmed and '\0' is assigned to the 12th byte.
You may wish to use the islower function to check is a character is lowercase. So you may change
if (a == 97)
to
if (islower(a)) // check if a character is lowercase.
Remember you may need to include the string.h header to use islower()