(C Programming) How do I insert string into a character array and then print out all the elements of this array?

Question

I have this string that is presented in the form of character array temp. And I want to insert this character array to another array temp_list and then print out the contents of this array. In other words, storing many character arrays in a single array. Can anyone tell me if this is possible and how can I make it work?

This is an example of what I am trying to accomplish:

int main()
{
    char temp[5] = "begin";
    char temp_list [10];
    temp_list[0] = temp;

    for (int i = 0; i < strlen(temp_list); i++)
    {
        printf("Labels: %s,", temp_list[i]);
    }
}

When I run this program, it prints out gibberish.

Any form of guidance would be very appreciated. Thank you.

Edit:

Thank you for the answers. They are all really helpful. But I have another question... what if I have multiple character arrays that I want to insert to temp_list? Using strcpy multiple times seem to not work, since I am assuming the function basically replaces the entire content of the temp_list with the string passed with strcpy?

You should enable all warnings and listen to them. temp_list[0] = temp; This is not how a string is copied. — Gerhardh
– Gerhardh, Commented Feb 25, 2019 at 15:33
Does that code compile? You have an assignment from char* to char which shouldn’t be ok. — Sami Kuhmonen
– Sami Kuhmonen, Commented Feb 25, 2019 at 15:33

Blaze · Accepted Answer · 2019-02-25 15:33:21Z

5

There are a bunch of misconceptions regarding strings. Your array temp needs to be big enough to also store the null-terminator, so it needs a size of at least 6 in this case:

char temp[6] = "begin"; // 5 chars plus the null terminator

To copy the string, use strcpy:

char temp_list[10];
strcpy(temp_list, temp);

To print it, pass temp_list, not temp_list[i], also you don't need that loop:

printf("%s\n", temp_list);

The final program could look like this:

int main()
{
    char temp[6] = "begin";
    char temp_list[10];
    strcpy(temp_list, temp);
    printf("%s\n", temp_list);
    return 0;
}

answered Feb 25, 2019 at 15:33

Blaze

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7 Comments

Tina Over a year ago

Thank you. But what if I would like to insert more than one character array? I guess, strcpy would no longer be an option, right? Since the function replaces the entire array with one string?

Blaze Over a year ago

Depends on how you want to insert more. strcpy copies the string, but doesn't change anything after the null-terminator. You can also use memcpy to copy a fixed amount of bytes, and strcat lets you concatenate strings. For instance if after strcpy(temp_list, temp); you would call strcat(temp_list, "Hi!"); it would print beginHi!

Tina Over a year ago

I was thinking to make it like {"hello", "world"} sort of array. Is that not possible in C?

Blaze Over a year ago

You can try: char temp[][10] = {"hello", "world"}; char temp_list[20]; strcpy(temp_list, temp[0]); strcat(temp_list, temp[1]);

Blaze Over a year ago

@Tina In this case you probably want temp_list to be a 2D array, too, like this:

char temp_list[2][20]; strcpy(temp_list[0], temp[0]); strcpy(temp_list[1], temp[1]); printf("%s %s\n", temp_list[0], temp_list[1]);

|

dbush · Accepted Answer · 2019-02-25 15:42:30Z

4

You have three problems here. First, temp is not big enough to hold the string "begin". Strings in C are null terminated, so this string actually takes up 6 bytes, not 5. So make temp big enough to hold this string:

char temp[6] = "begin";

Or better yet:

char temp[] = "begin";

Which sizes the array exactly as needed for the string. The second problem is here:

temp_list[0] = temp;

You're assigning an array (actually a pointer to the array's first element) to the first element of another array. That's a type mismatch of assigning a char * to a char. Even if the types matched, that's not how strings are copied. For that, use the strcpy function:

strcpy(temp_list, temp);

Finally, you're not printing the result correctly:

for (int i = 0; i < strlen(temp_list); i++)
{
    printf("Labels: %s,", temp_list[i]);
}

The %s format specifier expects a pointer to a char array in order to print a string, but you're passing in a single characters. Mismatching format specifiers invokes undefined behavior.

For printing single characters, use %c instead:

for (int i = 0; i < strlen(temp_list); i++)
{
    printf("Labels: %c,", temp_list[i]);
}

Or you can get rid of the loop and just print the whole string using %s:

printf("Labels: %s", temp_list);

edited Feb 25, 2019 at 15:42

answered Feb 25, 2019 at 15:35

dbush

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2 Comments

machine_1 Over a year ago

One more thing: he is using incorrect format specifier to print char in printf.

dbush Over a year ago

@machine_1 Good catch. Updated.

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