Strings and Characters in arrays.

Hi,

I am pretty new to C and it would be so great if I could get some help asap.

I am currently trying to read a string from user input, so I have:

printf("please enter your name\n");
scanf ("%s", &name);

But here's the part i'm lost at:
so say the user enters their name 'jane'
I would like to capture each letter separately as characters, in an array, so I can print it out like
j
a
n
e
I would also like to then know how I assign a letter a number, like a-z is 1-26. So that jane could alternatively be printed out as
10, 1, 14, 5

ive tried everything I can think of. Please help!

scanf ("%s", name);

that should solve your problem

and array can be passed as an address

I already had

scanf("%s", &name); for getting the input string.

So does taking away the & help? I still need to know how to get each letter captured as a single character, so i can do a for loop and print it like
j
a
n
e.

I am not great at explaining, sorry :confused:

like i tried this:

Code:

---

int main () {

char* name[30];
int i;
int length;

printf("please enter your name\n");
scanf ("%s", &name);
length = strlen(plaintext);
for(i=0;i<length;i++){
printf ("%c", plaintext[i]);  }

}

---

but it doesnt work, it outputs gobbledegook.

ty

Quote:

---

Originally Posted by akidamo

I already had

scanf("%s", &name); for getting the input string.

So does taking away the & help? I still need to know how to get each letter captured as a single character, so i can do a for loop and print it like
j
a
n
e.

I am not great at explaining, sorry :confused:

like i tried this:

Code:

---

int main () {

char* name[30];
int i;
int length;

printf("please enter your name\n");
scanf ("%s", &name);
length = strlen(plaintext);
for(i=0;i<length;i++){
printf ("%c", plaintext[i]);  }

}

---

but it doesnt work, it outputs gobbledegook.

ty

---

char* is a pointer not an array, char* name[30] is an array of pointers if you want to use a pointer then allocate memory for the pointer to point to so you can put data into the allocated memory, or use a character array.
plaintext, I don't see this variable decalred?
and lose the & like was already said previously.

FAQ > How do I... (Level 1) > Get a line of text from the user/keyboard (C)

ok yea, sorry plaintext was supposed to be name, i just changed it for the post, forgot to change that. Ok thanks, like i said im really new to this and the rules, ill try what you've suggested, thanks.

yea thanks it works :) so easy

I was being silly with the pointers thingy, I had been trying to teach myself C from online documentation, and I got mixed up.

But easy when u know how :)

Just one more question though.... to assign the letters a-z a numerical value, is there a quick way of doing this?

So I can have

10
1
14
5

as alternative output, when a user inputs a string?

Thanks

If you can assume ASCII or some other character set that guarantees that alphabetic characters are contiguous, then you could do something like this.

Code:

---

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

int main()
{
  char line[80];
  fputs("text? ", stdout);
  fflush(stdout);
  if ( fgets(line, sizeof line, stdin) != NULL )
  {
      int i;
      char *newline = strchr(line, '\n');
      if ( newline != NULL )
      {
        *newline = '\0';
      }
      for ( i = 0; line[i] != '\0'; ++i )
      {
        printf("%c %d\n", line[i], toupper(line[i]) - 'A' + 1);
      }
  }
  return 0;
}

/* my output
text? Jane
J 10
a 1
n 14
e 5
*/

---

Many thanks, thats been great help :)

Hi again,

sorry another silly question,
I now want to do a mathematical function, so I can put an int to its power of 3, then modulus 33.
eg: num2 = num1(to power of 3) % 33
But I dont know how to tell the program to put the int to Power of 3. Can you help?

Easy:
Multiply num1 3 times

Not as easy:
include math.h and use the pow() function

thanks yea ill have to use pow() then, cus its not always necessarily to power of 3, thanks :)

another problem sigh....with a different program

could I please again have some help?

I want the program to read in a list of integers to sort, from the user.
I then want to split this list in half. Say the list has 6 integers. I want to put the first 3 into a new array, and the last three into a third array.
I can get the first 3 in a second array by doing this:

Code:

---

split = n/2;

                  for (i = 0; i < split; i++) {
                          ab[i] = a[i];
                printf ("this is in ab %d\n", ab[i]);
                Odd_even_sort(ab, split);    }
        }

---

which is probably not very efficient, but anyway i am having trouble figuring out how to get the last three in the third array.

Me doh, sorry.

Code:

---

half = totalcount / 2;
for( x = 0; x < half; x++ )
    newarray1[ x ] = oldarray[ x ]; /* copy the first half */

for( y = 0, x = half; y < half; y++, x++ )
    newarray2[ y ] = oldarray[ x ]; /* copy the second half */

---

Or, we could abuse the way the first for loop works with x, and do this:

Code:

---

for( x = 0; x < half; x++ )
    newarray1[ x ] = oldarray[ x ];

for( y = 0; y < half; y++, x++ )
    newarray2[ y ] = oldarray[ x ];

---

We could further abuse x and y, but I'll leave that up to you.


Quzah.

Ok yesterday I was helped converting a letter to a number, like a to 1, b to 2, c to 3 etc... with the code in a previous post.

So I modified that to work in my own program.
but now I have the opposite problem, I would like to read in a small string of numbers, and convert them back to the letters.

say a person enters "10 1 5 6", ...then the computer would output jane.

Ive tried but failing miserably, theres a few problems i can think of:
getting the whole string, and ignoring the spaces.
getting the numbers separate,
damn i'm a bit lost where I start,
could anyone point me in the right direction.
Sorry!

Write a function to convert a letter to a number.
Write a function to convert a number to a letter.
Run your input through your function of choice.


Quzah.

ok
just while im in the middle of trying stuff then....
I noticed that when I have

temp = pow(array1[i],3);

it calculates the right number,

but if I have

temp = pow(array1[i],7);

it doesnt calculate the number to the power of 7 correctly. Does the pow() function not allow past a certain number?
Is there anyway I could make this work....

ok wait i have to use something larger than an int dont i?

ok i donno :(

Code:

---

int i;
double temp2;
double cseven[30];

                        printf("\n Decryption: \t");
                for (i=0; i<length; i++) {
                        temp2 =  pow(array1[i],7);
                        cseven[i] = temp2%33;
                        printf(" %f  ", cseven[i]);

---

}

when I try longints, it doesnt work, when I try floats or doubles I know it works because if I print out temp2 afterwords it is correct. But if I then do: cseven[i] = temp2%33; etc... as above when compiling I get the error:

illegal use of floating point in function
temp2 is assigned a value that is never used...

please help

Perhaps you should convert the double into an int and then do it Not sure it'll work but it's just an idea.

nope now im gettin the wrong answer again :(

It seems you cant modulus a float or double,
and when i do this....

Code:

---

int i;
        double temp2;
        int cseven[30];
        long temp3;


                  printf("\n Decryption: \n");
                                for (i=0; i<length; i++) {
                          temp2 = pow(array1[i],7);
                        temp3 = temp2;
                        printf(" %ld \n ", temp3);
                                                printf(" %ld \n ", temp3%33);
                        }

---

It seems to only work to an extent, if array1[i] is like over a certain number, it won't work. It works if array1[i] is 21, but not 26.
so stuck.....

26^7 is what number? Big. Too big in fact to fit in a 32 bit number. You can't hold infinity in 32 bits.


Quzah.

Yea i figured size was the problem, The answer for 26^7 is 8031810176. I just wanted to know was there a way to get around it.

ok figured out how to get the right answer,

after using alot of temps!

Code:

---

printf("\n Decryption: \n");
                          for (i=0; i<length; i++) {
                          temp2 = (double)pow(ciphertext[i],7);
                                temp3 = temp2/33;
                                temp5 = temp3;
                                temp6 = temp5;
                                temp4 = temp6*33;
                                printf(" temp 5 is %ld \n\n ", temp5);
                                printf(" temp 4 is %f \n ", temp4);
                                temp7 = temp2-temp4;
                                cseven[i] = temp7;
                                printf(" temp 7 is %d \n ", cseven[i]);
                        }

---

It prints the correct answers as would using %. Thanks for all help given.