C - Print ASCII Value for Each Character in a String

Question

I'm new to C and I'm trying to write a program that prints the ASCII value for every letter in a name that the user enters. I attempted to store the letters in an array and try to print each ASCII value and letter of the name separately but, for some reason, it only prints the value of the first letter.

For example, if I write "Anna" it just prints 65 and not the values for the other letters in the name. I think it has something to do with my sizeof(name)/sizeof(char) part of the for loop, because when I print it separately, it only prints out 1.

I can't figure out how to fix it:

#include <stdio.h>

int main(){

    int e;
    char name[] = "";
    printf("Enter a name : \n");
    scanf("%c",&name);
    for(int i = 0; i < (sizeof(name)/sizeof(char)); i++){
        e = name[i];
        printf("The ASCII value of the letter %c is : %d \n",name[i],e);
    }
    int n = (sizeof(name)/sizeof(char));
    printf("%d", n);
}

Answer 1

If we need write a code to get ASCII values of all elements in a string, then we need to use "%d" instead of "%c". By doing this %d takes the corresponding ascii value of the following character. If we need to only print the ascii value of each character in the string. Then this code will work:

#include <stdio.h>
char str[100];
int x;
int main(){
    scanf("%s",str);
    for(x=0;str[x]!='\0';x++){
        printf("%d\n",str[x]);
    }
}

To store all corresponding ASCII value of character in a new variable, we need to declare an integer variable and assign it to character. By this way the integer variable stores ascii value of character. The code is:

#include <stdio.h>
char str[100];
int x,ascii;
int main(){
    scanf("%s",str);
    for(x=0;str[x]!='\0';x++){
        ascii=str[x];
        printf("%d\n",ascii);
        
    }
}

I hope this answer helped you.....😊

Answer 2

This is a rather long answer, feel free to skip to the end for the code example.

---

First of all, by initialising a char array with unspecified length, you are making that array have length 1 (it only contains the empty string). The key issue here is that arrays in C are fixed size, so name will not grow larger.

Second, the format specifier %c causes scanf to only ever read one byte. This means that even if you had made a larger array, you would only be reading one byte to it anyway.

The parameter you're giving to scanf is erroneous, but accidentally works - you're passing a pointer to an array when it expects a pointer to char. It works because the pointer to the array points at the first element of the array. Luckily this is an easy fix, an array of a type can be passed to a function expecting a pointer to that type - it is said to "decay" to a pointer. So you could just pass name instead.

As a result of these two actions, you now have a situation where name is of length 1, and you have read exactly one byte into it. The next issue is sizeof(name)/sizeof(char) - this will always equal 1 in your program. sizeof char is defined to always equal 1, so using it as a divisor causes no effect, and we already know sizeof name is equal to 1. This means your for loop will only ever read one byte from the array. For the exact same reason n is equal to 1. This is not erroneous per se, it's just probably not what you expected.

---

The solution to this can be done in a couple of ways, but I'll show one. First of all, you don't want to initialize name as you do, because it always creates an array of size 1. Instead you want to manually specify a larger size for the array, for instance 100 bytes (of which the last one will be dedicated to the terminating null byte).

char name[100];
/* You might want to zero out the array too by eg. using memset. It's not
   necessary in this case, but arrays are allowed to contain anything unless
   and until you replace their contents.

   Parameters are target, byte to fill it with, and amount of bytes to fill */
memset(name, 0, sizeof(name));

Second, you don't necessarily want to use scanf at all if you're reading just a byte string from standard input instead of a more complex formatted string. You could eg. use fgets to read an entire line from standard input, though that also includes the newline character, which we'll have to strip.

/* The parameters are target to write to, bytes to write, and file to read from.

   fgets writes a null terminator automatically after the string, so we will
   read at most sizeof(name) - 1 bytes.
*/
fgets(name, sizeof(name), stdin);

Now you've read the name to memory. But the size of name the array hasn't changed, so if you used the rest of the code as is you would get a lot of messages saying The ASCII value of the letter is : 0. To get the meaningful length of the string, we'll use strlen.

NOTE: strlen is generally unsafe to use on arbitrary strings that might not be properly null-terminated as it will keep reading until it finds a zero byte, but we only get a portable bounds-checked version strnlen_s in C11. In this case we also know that the string is null-terminated, because fgets deals with that.

/* size_t is a large, unsigned integer type big enough to contain the
   theoretical maximum size of an object, so size functions often return
   size_t.

   strlen counts the amount of bytes before the first null (0) byte */
size_t n = strlen(name);

Now that we have the length of the string, we can check if the last byte is the newline character, and remove it if so.

/* Assuming every line ends with a newline, we can simply zero out the last
   byte if it's '\n' */
if (name[n - 1] == '\n') {
    name[n - 1] = '\0';
    /* The string is now 1 byte shorter, because we removed the newline.
       We don't need to calculate strlen again, we can just do it manually. */
    --n;
}

The loop looks quite similar, as it was mostly fine to begin with. Mostly, we want to avoid issues that can arise from comparing a signed int and an unsigned size_t, so we'll also make i be type size_t.

for (size_t i = 0; i < n; i++) {
    int e = name[i];
    printf("The ASCII value of the letter %c is : %d \n", name[i], e);
}

Putting it all together, we get

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

int main() {

    char name[100];
    memset(name, 0, sizeof(name));
    printf("Enter a name : \n");
    fgets(name, sizeof(name), stdin);

    size_t n = strlen(name);
    if (n > 0 && name[n - 1] == '\n') {
        name[n - 1] = '\0';

        --n;
    }

    for (size_t i = 0; i < n; i++){
        int e = name[i];
        printf("The ASCII value of the letter %c is : %d \n", name[i], e);
    }


    /* To correctly print a size_t, use %zu */
    printf("%zu\n", n);

    /* In C99 main implicitly returns 0 if you don't add a return value
       yourself, but it's a good habit to remember to return from functions. */
    return 0;
}

Which should work pretty much as expected.

---

Additional notes:

• This code should be valid C99, but I believe it's not valid C89. If you need to write to the older standard, there are several things you need to do differently. Fortunately, your compiler should warn you about those issues if you tell it which standard you want to use. C99 is probably the default these days, but older code still exists.

• It's a bit inflexible to be reading strings into fixed-size buffers like this, so in a real situation you might want to have a way of dynamically increasing the size of the buffer as necessary. This will probably require you to use C's manual memory management functionality like malloc and realloc, which aren't particularly difficult but take greater care to avoid issues like memory leaks.

• It's not guaranteed the strings you're reading are in any specific encoding, and C strings aren't really ideal for handling text that isn't encoded in a single-byte encoding. There is support for "wide character strings" but probably more often you'll be handling char strings containing UTF-8 where a single codepoint might be multiple bytes, and might not even represent an individual letter as such. In a more general-purpose program, you should keep this in mind.

Answer 3

Here's a corrected, annotated version:

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

int main() {
    int e;
    char name[100] = ""; // Allow for up to 100 characters
    printf("Enter a name : \n");
//  scanf("%c", &name); // %c reads a single character
    scanf("%99s", name); // Use %s to read a string! %99s to limit input size! 
//  for (int i = 0; i < (sizeof(name) / sizeof(char)); i++) { // sizeof(name) / sizeof(char) is a fixed value!
    size_t len = strlen(name); // Use this library function to get string length
    for (size_t i = 0; i < len; i++) { // Saves calculating each time!
        e = name[i];
        printf("The ASCII value of the letter %c is : %d \n", name[i], e);
    }
    printf("\n Name length = %zu\n", strlen(name)); // Given length!
    int n = (sizeof(name) / sizeof(char)); // As noted above, this will be ...
    printf("%d", n);                 // ... a fixed value (100, as it stands).
    return 0; // ALWAYS return an integer from main!
}

But also read the comments given in your question!