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pointers on arrays

This is a discussion on pointers on arrays within the C Programming forums, part of the General Programming Boards category; Code: #include <stdio.h> int main(){ char *str; char t; int i=0, j=0; printf("\n"); printf("Enter String (up to 50 chars): "); ...

  1. #1
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    pointers on arrays

    Code:
    #include <stdio.h>
    int main(){
    char *str;
    char t;
    int i=0, j=0;
    printf("\n");
    printf("Enter String (up to 50 chars): ");
    scanf("%s", *str[j]);
    	for (j = 0; *str; ++j);
    
    	i==j;
    	j==0;
    
    	while(i>=j){
    	t = *str[i];
    	*str[i] = *str[j];
    	*str[j] = t;
    	i--;
    	j++;
    
    	}
    
    	printf("%s\n",(str));
    	
    
    printf("\n");
    }

    what i'm i doing wrong...

  2. #2
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    Many, many things:
    • main returns an int, but you fail to do so.
    • char *str, is a pointer that you never actually allocated any space for, however you attempt to store items in it
    • You seem to be confusing a char array and a char in your loops
    • Your compiler should be having a fit over this code, I suggest you pay attention to it or get a new compiler

    You will benefit from reading through our C Tutorials. Yes, all of them; there are serious conceptual flaws with this code.
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    ..... Just don't be surprised when I say you aren't using standard C anymore, and as such,are off in your own little universe that I will completely disregard.
    Warning: Some or all of my posted code may be non-standard and as such should not be used and in no case looked at.

  3. #3
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    What are u basically up to??? I don't understand where to start from....Okay lets give it a try to explain ur mistakes.....

    Code:
    char *str;
    str is a character pointer that can hold only 1 address which simply points to a character or group of characters(string). In this case, it holds the starting address of the input string.

    str is not an array. So *str[j] is not correct but str[j] may be correct. You can use this to swap characters in the string.

    Try,
    Code:
    scanf("%s",str);
    instead of
    Code:
    scanf("%s",*str[j]);
    Why do u give *str in the for loop's condition? *str means the 1st character of the input string, when its correctly formatted; otherwise its raw bits which makes no sense to the for loop. It probably crashes ur program.

    U can find the length of the string by strlen(str)

    == is a check for equality which evaluates to either 0 or 1. I think its =, that's what u need or searching for, may be.

    Again, the following is wrong as str is not an array. [You confused between pointer to a string and an array of null-terminated characters(string).]
    Code:
    t = *str[i];
    *str[i] = *str[j];
    *str[j] = t;
    *str[i] makes no sense.

    Dont get discouraged....u simply need a to read ur text book properly!!!
    Last edited by Avenger625; 09-11-2011 at 12:07 PM.

  4. #4
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by Avenger625
    str is not an array. So str[j] is not correct.
    It is true that str is not an array, but depending on the value of j, str[j] may be correct. str[j] would be equivalent to *(str + j).
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  5. #5
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    Right!!!

  6. #6
    Algorithm Dissector iMalc's Avatar
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    Quote Originally Posted by antros48 View Post
    what i'm i doing wrong...
    Not telling us what the problem is, and not asking a question about how to fix it.
    My homepage
    Advice: Take only as directed - If symptoms persist, please see your debugger

    Linus Torvalds: "But it clearly is the only right way. The fact that everybody else does it some other way only means that they are wrong"

  7. #7
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    Quote Originally Posted by antros48 View Post
    what i'm i doing wrong...
    Although nobody has come straight out and said it... You need to allocate memory for your input string before you do anything else...
    You have two chioices here...
    Code:
    // choice #1
    char str[51];  
    
    //choice #2
    char *str = malloc (51 * sizeof(char));
    If you choose #2, you need to free() the allocated memory when you are done with it.
    Avenger625 likes this.

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