Changing the value of a character in an array of strings in C

Question

I am trying to figure out how to change the letter "j" to a "Y" using a pointer in the program below:

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

int main()
{

    char *buffer[] = {"ABCDEFGH", "ijklmnop", "QRSTUVWX"};

    printf("the value as %%s of buffer[0] is %s\n", buffer[0]);
    printf("the value as %%s of buffer[1] is %s\n", buffer[1]);
    printf("the value as %%s of buffer[2] is %s\n", buffer[2]);
    printf("the sizeof(buffer[2]) is %d\n", sizeof(buffer[2]));


    printf("the value as %%c of buffer[1][3] is %c\n", buffer[1][3]);
    printf("the value as %%c of buffer[2][3] is %c\n", buffer[2][3]);

    /*create a pointer to the pointer at buffer[1]*/
    char *second = buffer[1];

    /*change character at position 2 of bufffer[1] to a Y*/
    second++;
    second = "Y";

    printf("character at location in second is %c\n", *second);
    printf("character at location in second+2 is %c\n", second+2);

    printf("the values as %%c of second second+1 second+2 second+3 are %c %c %c %c\n",*(second),*(second+1),*(second+2),*(second+3));

    return(0);

}

How can I change a character located in an array of strings?

Trying to change string literals like that is undefined behaviour. — paxdiablo
– paxdiablo, Commented May 16, 2013 at 21:54

Paul R · Accepted Answer · 2013-05-16 21:54:27Z

2

Attempting to write to string literals is undefined behaviour.

Change:

char *buffer[] = {"ABCDEFGH", "ijklmnop", "QRSTUVWX"};

to:

char buffer[][10] = {"ABCDEFGH", "ijklmnop", "QRSTUVWX"};

answered May 16, 2013 at 21:54

Paul R

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Comments

Sam I am says Reinstate Monica · Accepted Answer · 2013-05-16 21:54:57Z

2

Just a thing I noticed

char *second = buffer[1];

/*change character at position 2 of bufffer[1] to a Y*/
second++;
second = "Y";

You're trying to set to the string "Y" instead of to the char 'Y'.

I'd do something more like

*second = 'Y';

answered May 16, 2013 at 21:54

Sam I am says Reinstate Monica

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Comments

Nik Bougalis · Accepted Answer · 2013-05-16 22:09:03Z

1

There's two things that are going to make this not work:

First, let's look at how you modify things:

second++;
second = "Y";

Think about what you are doing here: you are making second point to the string "Y". This is certainly not your intent. You want to change the second character of whatever string second points to... there are two ways to go about that:

/* make the second character of whatever string the variable second to into a 'Y' */
second++;
*second = 'Y';

Alternatively, you could use the more intuitive:

second[1] = 'Y';

Now, even with this fix, your code will attempt to modify string literals, which will result in undefined behavior. Consider your code:

char *buffer[] = {"ABCDEFGH", "ijklmnop", "QRSTUVWX"};

At this point, buffer[0], buffer[1] and buffer[2] point to memory that (may) be read-only. Whether it actually is or not is irrelevant though; you should certainly TREAT it as read-only. Therefore, the attempt to change anything would fail. To avoid that issue change things slightly:

char *buffer[3];

buffer[0] = strdup("ABCDEFGH");
buffer[1] = strdup("ijklmnop");
buffer[2] = strdup("QRSTUVWX");

And it goes without saying, but be sure to call free to release the strings pointed to by buffer[0], buffer[1] and buffer[2] when you're done with them!

edited May 16, 2013 at 22:09

answered May 16, 2013 at 22:02

Nik Bougalis

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1 Comment

Nik Bougalis Over a year ago

Man... everyone else types so fast!

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