array of N pointers

This is a discussion on array of N pointers within the C Programming forums, part of the General Programming Boards category; Hello, I need some help on this: Isn't int *array[3]; the same thing as int *array = (int*) malloc(3*sizeof(int*)); ?? ...

  1. #1
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    array of N pointers

    Hello, I need some help on this:

    Isn't

    int *array[3];

    the same thing as

    int *array = (int*) malloc(3*sizeof(int*)); ??

    I used the first declaration to print the integers: *array[i] and it worked.

    When I used the second one for N pointers, I get an error about invalid argument type of 'unary*' (have 'int').

    Thank you.

    Edit:

    Maybe pasting the whole code it will help:

    Code:
    #include <pthread.h>
    #include <stdlib.h>
    #include <stdio.h>
    #include <sys/types.h>
    #include <unistd.h>
    #include <errno.h>
    
    void *myfunc(void *arg)
    {
    	int *x = (int*) arg;
    	printf("O meu pid  %u, o meu tid  %u e recebi arg=%d \n", (unsigned) getpid(), (unsigned) pthread_self(), *x);	               
    	return (void*) x;
    }
    
    int main (int argc, char* argv[])
    {
    	if (argc != 2)
    	{
    		perror("Falta de argumentos!\n");
    		printf("Erro %d\n", errno);
    		_exit(1);
    	}	 
    	int n = atoi(argv[1]), st, i; 
    	pthread_t *tids = (pthread_t*) malloc(n*sizeof(pthread_t));
    	int *retval = (int*) malloc(n*sizeof(int*));
    	for(i=1; i<=n; i++)
    	{	 
    		st= pthread_create(&tids[i-1], NULL, myfunc, (void*) &i);
    		if (st != 0) 
    		{ 
    			perror("Erro em pthread_create()\n"); 
    			printf("Erro %d\n", errno);
    			_exit(2); 
    		}
    	}
    	
    	for (i=1; i<=n; i++)
    	{
    		st = pthread_join(tids[i-1], (void**) &retval[i-1]); 
    		if (st != 0) 
    		{ 
    			perror ("Erro em pthread_join()\n"); 
    			printf("Erro %d\n", errno);
    			_exit(3);
    		}
    		printf("Thread %u devolveu %d\n", (unsigned) tids[i-1], *retval[i-1]);
    	}
    	_exit(0);
    }
    Last edited by Lima; 06-06-2009 at 11:24 AM.

  2. #2
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    int *array = (int*) malloc(3*sizeof(int*)); should be


    int *array = (int*) malloc(3*sizeof(int));

    You want three integer cells, not three integer pointer cells. I think.

  3. #3
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    Quote Originally Posted by CLimaPT
    Isn't

    int *array[3];

    the same thing as

    int *array = (int*) malloc(3*sizeof(int*)); ??
    No, they are not the same. The former declares an array of 3 pointers to int; the latter declares a pointer to int.

    If you want a dynamic array of 3 pointers to int, you should write:
    Code:
    int **array = malloc(3 * sizeof(*array));
    Quote Originally Posted by CLimaPT
    When I used the second one for N pointers, I get an error about invalid argument type of 'unary*' (have 'int').
    What is the exact error message?
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  4. #4
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    I get this:

    f3exercicio1.c: In function ‘main’:
    f3exercicio1.c:36: error: invalid type argument of ‘unary *’ (have ‘int’)

    Oh, I think I get it, thank you once again. No cast needed by the way?
    Last edited by Lima; 06-06-2009 at 11:47 AM.

  5. #5
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by CLimaPT
    f3exercicio1.c: In function ‘main’:
    f3exercicio1.c:36: error: invalid type argument of ‘unary *’ (have ‘int’)
    Line 36 appears to be a blank line... so which line is line 36?

    By the way, there is no need to cast a pointer to be a pointer to void. The conversion will happen implicitly. There is also no need to cast a void* to be the desired pointer type unless you are trying to be compatible with C++.

    You should also get used to the idiom of starting your loop counters from 0 instead of 1 when you are using the loop counters to index an array.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  6. #6
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    This line:

    printf("Thread %u devolveu %d\n", (unsigned) tids[i-1], *retval[i-1]);

    It was line 46, my bad, pasted the other program output.

  7. #7
    C++ Witch laserlight's Avatar
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    Right. Since retval is a pointer to int, retval[i-1] is an int, so *retval[i-1] does not make sense. Perhaps you mean to make retval a pointer to pointer to int, as I demonstrated in post #3.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  8. #8
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    Yeah, I see my mistake now. :P

    Thanks.

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