While loop to count the charecters of a string

Hello,

What's wrong with:

Code:

---

int getLength(char* A){
        int i = 0;
        while (*(A+i) != '/0') i++;
        return i;

---

I get an infinite loop at the while statement.

Seems unlikely. What's the rest of the code?

--
Mats

Code:

---

char A[] = "Alice was beginning to get tired of sitting by her sister on the bank, and of having nothing to do.";

int getLength(char* A){
        int i = 0;
        while (*(A+i) != '/0') i++;
        return i;
}

int main(){
        printf("%d\n",getLength(*A));
        return 0;

---

Sorry, missed that you are using / instead of \ in what should be '\0'.

--
Mats

What about this:

Code:

---

void wrap(char* A, int m){
        char* B = A;
        int n = getLength(B);
        int j,i;
        for (i=m; i<n; i+=m){
                j = i+1;
                while (&(A+j) != " ") j--;  //error
                &(A+j) = "\n"; //error
        }
}

---

actually there should be * follow the pointer, but that gives a warning too.

You are using double quotes when comparing to a char. A char is in single quotes.

why to mix pointer notation when you have index?

Code:

---

A[j] != ' '

---

is a lot easier to read

I'm wondering why this code gives bad access signal:
To me it should copy an array passed as A.

Code:

---


void costCalc(char* A, int m){
        char* B;
        int i=0;
        /*while (*(A+i) != '\0'){ *(B+i) = *(A+i); i++;};
        *(B+i) = '\0';

---

This thread, ladies and gentlemen, is why you should compile with warnings on.

Quzah.

Why this:

Code:

---


void wrapper(char A[],int n, int m){
        int i= 0;
        while (A[i] != ' ' && A[i] != '\0' && i<n) i++; //so from 0 to i we have the first word assuming we never start with a space, in case of "aa" returns 2
        A[i] = '\n';

---

gives bad access signal when passed "aa" as char and n as 3?

How are you calling the function - most likely, you are writing to a place in memory that isn't writable.

--
Mats