C string question

Hi, I recently started learning programming in C, I am not a programming newbie, just new to C, now here is my question:

I am experimenting with strings, because they work in an entirely different way from what I am used to, this is what i did

Code:

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int teller;
int aantal = 3;
int getal[aantal];
int *ptr;

ptr = &getal[0];
*ptr = 1;
ptr = &getal[1];
*ptr = 2;
ptr = &getal[2];
*ptr = 3;

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this is not my problem, the problem is when I try to output the string, i found a way around the problem using this:

Code:

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for (teller = 0; teller <= (aantal-1); teller++)
{
ptr = &getal[teller];
printf("%d",*ptr);
}

---

but I believer there is a better way to output it, but I can't seem to find it, I always get an error during compiling like "format '%s expects type 'char *' but argument 2 has type 'int *'

any help?

That's because a string is a null-terminated array of type char. But you have a simple (not null-terminated) array of type int.

The only string in all of your code is:

Code:

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"%d"

---

You can simplify the loop somewhat like this -- incremting the pointer on each loop interation instead of reseting it to some other value.

Code:

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int* ptr = getal;
for (teller = 0; teller < aantal; teller++)
{
  printf("%d",*ptr++);
}

---

ok, it's not a string, but still, if I try to ouput "geheel" the normal way:

Code:

---

printf("%d",getal);

---

the compiler says:
"format '%d expects type 'int' but argument 2 has type 'int *'"

your compiler is right! see my previos post for the right solution. You can't use character array solution to output int arrays.

My original idea was trying to modify one element of a character string, but since that didn't seem to work out, I first tried this, now, then, can you tell me how to change an element of a string?

f.e. if i do this

Code:

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char word[3] = "foo";

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and I wanted to change the "f" into a "b", how do I do this?

way #1:

Code:

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word[0] = 'b';

---

way #2

Code:

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*word = 'b';

---

Please note though that your declaration is incorrect. You're making an array of 3 elements but storing 4 characters in it. "foo" in memory is actually 'f', 'o', 'o', '\0'. Every string in C must end with the value 0.

So your char word[3] = "foo"; is actually equivalent to char word[3] = { 'f', 'o', 'o', '\0' }; See the problem?

yes, i see the problem, it's a common made mistake for c newbies, I've read it on several sites, I hope I won't forget the null terminator in the future

hmm, this shows that I've got much to learn, I tried that, but always used " instead of '

thanks for the help

define it as below and the compiler will figure it out for you. That way you can create very large strings without counting up the number of characters -- compilers are good at doing that.

Code:

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char word[] = "foo";

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