In this line:

scanf("%s", &word[i]);

You need to make sure word[i] is pointing somewhere, and has enough space to occupy the string entered. Since word[i] is a char * pointer, you need to at some time allocate memory for this. Otherwise, it is just a dangling pointer not pointing anywhere.

If you want to stick with scanf(), then you can allocate some space beforehand with malloc.

malloc() allocates requested memory on the heap, then returns a void* pointer at the end.

You can apply malloc() in your code like this:

size_t malloc_size = 100;

for (i = 0; i < 3; i++) {
    word[i] = malloc(malloc_size * sizeof(char)); /* allocates 100 bytes */
    printf("Enter word: ");
    scanf("%99s", word[i]); /* Use %99s to avoid overflow */
                            /* No need to include & address, since word[i] is already a char* pointer */
} 

Note: Must check return value of malloc(), because it can return NULL when unsuccessful.

Additionally, whenever you allocate memory with the use of malloc(), you must use free to deallocate requested memory at the end:

free(word[i]);
word[i] = NULL; /* safe to make sure pointer is no longer pointing anywhere */

Another approach without scanf

A more proper way to read strings should be with fgets.

char *fgets(char *str, int n, FILE *stream) reads a line from an input stream, and copies the bytes over to char *str, which must be given a size of n bytes as a threshold of space it can occupy.

Things to note about fgets:

• Appends \n character at the end of buffer. Can be removed easily.
• On error, returns NULL. If no characters are read, still returns NULL at the end.
• Buffer must be statically declared with a given size n.
• Reads specified stream. Either from stdin or FILE *.

Here is an example of how it can be used to read a line of input from stdin:

char buffer[100]; /* statically declared buffer */

printf("Enter a string: ");
fgets(buffer, 100, stdin); /* read line of input into buffer. Needs error checking */

Example code with comments:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define NUMSTR 3
#define BUFFSIZE 100

int main(void) {
    char *words[NUMSTR];
    char buffer[BUFFSIZE];
    size_t i, count = 0, slen; /* can replace size_t with int if you prefer */

    /* loops only for three input strings */
    for (i = 0; i < NUMSTR; i++) {

        /* read input of one string, with error checking */
        printf("Enter a word: ");
        if (fgets(buffer, BUFFSIZE, stdin) == NULL) {
            fprintf(stderr, "Error reading string into buffer.\n");
            exit(EXIT_FAILURE);
        }

        /* removing newline from buffer, along with checking for overflow from buffer */
        slen = strlen(buffer);
        if (slen > 0) {
            if (buffer[slen-1] == '\n') {
                buffer[slen-1] = '\0';
            } else {
                printf("Exceeded buffer length of %d.\n", BUFFSIZE);
                exit(EXIT_FAILURE);
            }
        } 

        /* checking if nothing was entered */
        if (!*buffer) {
            printf("No string entered.\n");
            exit(EXIT_FAILURE);
        }

        /* allocate space for `words[i]` and null terminator */
        words[count] = malloc(strlen(buffer)+1);

        /* checking return of malloc, very good to do this */
        if (!words[count]) {
            printf("Cannot allocate memory for string.\n");
            exit(EXIT_FAILURE);
        }

        /* if everything is fine, copy over into your array of pointers */
        strcpy(words[count], buffer);

        /* increment count, ready for next space in array */
        count++;
    }  

    /* reading input is finished, now time to print and free the strings */
    printf("\nYour strings:\n");
    for (i = 0; i < count; i++) {
        printf("words[%zu] = %s\n", i, words[i]);
        free(words[i]);
        words[i] = NULL;
    }

    return 0;
}

Example input:

Enter a word: Hello
Enter a word: World
Enter a word: Woohoo

Output:

Your strings:
words[0] = Hello
words[1] = World
words[2] = Woohoo

As mentioned in comments, you must change this:

/* bad */
int score [n];
printf("Number of scores: ");
scanf("%d", &n);

into this

/* good */
printf("Number of scores: ");
scanf("%d", &n);
int score [n];

This since C executes code from top to bottom like when you are reading a book. It will not "double back" a few rows above and fill in n once it has been entered by the user. At the point where you declare int score [n], n must already be known.

That should read scanf("%s",names[i]);

Right now, you are storing it as scanf("%s",names);, which is equivalent to scanf("%s",names[0]);

So, you are overwriting that same array entry in every pass.

EDIT: Also, when you pass char names[][] to a function, it only passes the pointer to the first element. You should declare atleast one bound of it, to the same value as the one to which you declared it.

int main(){
    //To accept 2 names of 2 characters each
    char names[2][2];// or char** names;
    initialize(names, 2,2);
}
void initialize(char names[][2],const int MAX_NAMES,const int MAX_NAMELENGTH){ .. }
                             ^ syntax error if index not present

(Reference)

Scanf() only accepts input for a single word.(not after space.)

To enter multiple words, you can use gets(string) or scanf("%[^\n]", string).

You can use scanf() function in order to achieve this.

The declaration of scanf function is as follows:

int scanf(const char *format, ...)

This means that while using scanf() function, you can specify the format of input you are receiving. Similar to printf() function, scanf("%c",&character), for example can be used to get character input from user. scanf() function returns the number of successful reads.

Note: You can check whether scanf() function successfully read the intended data by stating for example:

if (scanf("%d",&intNum)==1){
    //all is ok, proceed
}
else{
    //EOF or conversion failure
}

Likewise, if you are scanning for multiple variables, you can use if(scanf("%d %d",&intNum1,&intNum2)==2). Instead of %c, you can use any of the following(and more) in order to read and store different types of input.

Some of those:

c : character
d : decimal integer
f : float
o : octal integer
s : string of characters
etc...

Furthermore; for example if you use scanf("%d %f",&integerNum,&floatNum);, you will be reading user's input in format int float(int SPACE float) and storing the values into integerNum and floatNum variables respectively. The SPACE character between "%d %f" is part of the format.

The second argument (&character in my case) is the address scanf() is going to store the value into. For more information about scanf() : scanf(). You can use scanf() inside a loop to store user input into your array.

Here is a working sample:

#include <stdio.h>

int main(){
    int array[20];
    int i;
    printf("Enter value to be stored in array> ");
    for (i=0;i<20;i++){
        //printf("Enter value to be stored in array[%d]> ",i);
        scanf("%d,",&(array[i]));
        //scanf("%d,",(array+i));
    }
    for (i=0;i<20;i++){
        if (i==19){
            printf("%d\n",array[i]);
            //printf("%d\n",*(array+i));
        }
        else{
            printf("%d, ",array[i]);
        }
    }
    return 0;
}

In this example I used scanf("%d,",&(array[i])); since the format of the user input is going to be like this: int,int,int,int,...(int COMMA int COMMA int COMMA ...).

Notice how array returns the address of the first element of the array. By using pointers, you could rewrite scanf("%d,",&(array[i])); as scanf("%d,",(array+i));.

You can use this code to store other kinds of data like float with minimal changes. For example, in order to use this to get float input from user;

1. Replace int array[20] with float array[20]
2. Replace scanf("%d,",&(array[i])); with scanf("%f,",&(array[i]));

Also do not forget to use the right formatting when printing the values in your array.

Yes, but you need to make sure you have enough space to do so:

char input[3][50]; // enough space for 3 strings with
                   // a length of 50 (including \0)

fgets(&input[0], 50, stdin);

printf("Inputted string: %s\n", input[0]);

Using char **input does not have any space allocated for the input, therefore you cannot do it.

This is your immediate problem:

scanf("\n %c", str1);

This format string tells scanf to read some amount of whitespace followed by a single character, which is to be stored in str1.

@H2CO3 is right: scanf is almost never the right answer for reading input from the user. It performs no bounds checking and is too finicky about the format of user input; one unexpected character will completely confuse it. You are very strongly encouraged to use fgets instead if at all possible:

fgets(str1, 100, stdin);

If you are not allowed to do this, you should ask your professor why not. Seriously.

If you absolutely must use scanf to read input here, you can do it the way @user2479209 described in their answer.

There is another problem with your program, which is that toupper(chars[i]) will not change the value stored in chars[i]. You have to assign the result of toupper back to the array explicitly, e.g.:

chars[i] = toupper(chars[i]);

Some dialects of C have variable-length arrays (VLA), notably C99 (and GCC accepts them....)

But your teacher probably wants you to understand C dynamic memory allocation, and you should also test when scanf fails, so you might code:

int main(void) {
    int size, i;
    printf("Enter the size of the arrays:\n");
    if (scanf("%d", &size)<1) {
      perror("scan size");
      exit(EXIT_FAILURE);
    };

^{Actually, you'll better test against the case of size being negative! I leave that up to you.}

    int *arr1 = calloc(size, sizeof(int));
    if (arr1==NULL) {
       perror("calloc arr1");
       exit(EXIT_FAILURE);
    }

^{You always need to handle failure of calloc or malloc}

 printf("Enter the elements of the array:\n");
 for (i = 0; i < size; i++) {
    if(scanf("%d", &arr1[size])<1) {
      perror("scanf arr1[size]");
      exit(EXIT_FAILURE);
    }
 }

^{scanf_s is only existing in C11, and on a C11 compliant implementation you could use VLAs}

I leave the rest up to you. You want to code a for loop to print every element of the array.

Of course you should call free appropriately (and I am leaving that to you). Otherwise, you have a memory leak. Some systems have valgrind to help hunting them. Don't forget to compile with all warnings & debug info (with GCC use gcc -Wall -g) then use the debugger (e.g. gdb).

Don't forget that a programming language is a specification (written in some document), not a software. You may want to glance into n1570. And you need to understand what undefined behavior means.

For the %c specifier, scanf needs the address of the location into which the character is to be stored, but printf needs the value of the character, not its address. In C, an array decays into a pointer to the first element of the array when referenced. So, the scanf is being passed the address of the first element of the name array, which is where the character will be stored; however, the printf is also being passed the address, which is wrong. The printf should be like this:

printf("%c", name[0]);

Note that the scanf argument is technically ok, it is a little weird to be passing an array, when a pointer to a single character would suffice. It would be better to declare a single character and pass its address explicitly:

char c;
scanf("%c", &c);
printf("%c", c);

On the other hand, if you were trying to read a string instead of a single character, then you should be using %s instead of %c.