Cant find the syntax error....and clarification on what my program is doing.

gcc says: In function 'main':
error: syntax error before "int"
error: syntax error before "char"

Code:

#include <stdio.h>

int main(){
    int i;
    
    char char_array[5] = {'a', 'b', 'c', 'd', 'e'};
    int int_array[5] = {1,2,3,4,5};
    
    char *char_pointer;
    int *int_pointer;
    
    char_pointer = (char*) int_array; /*typecast into the pointers data type*/
    int_pointer = (int*) char_array;
    
    for(i=0; i < 5; i++){
        printf("[integer pointer] points to %p, which is contains the
        char '%c'\n", int_pointer, *int_pointer);
    int_pointer = (*int) ((char*)int_pointer + 1);
    }
    
    for(i=0; i < 5; i++){
        printf("[char pointer] points to %p, which contains the int %d\n",
        char_pointer, *char_pointer);
    char_pointer = (*char) ((int*)char_pointer + 1);
    }
}

i am following a book and the books doesnt really explain: char_pointer = (*char) ((int*)char_pointer + 1);
i would like to know what that piece of code is doing. also because ive already typecast in the beginning. so i dont see why its repeated.

EDIT: I also dont understand the argument. isnt typecasting supposed to have this format: (typecast_data) variable ? so wouldnt it be just ((int*) char_pointer+1); ?

Change those to these..

Originally Posted by prafiate

Code:

#include <stdio.h>

int main(){
    int i;
    
    char char_array[5] = {'a', 'b', 'c', 'd', 'e'};
    int int_array[5] = {1,2,3,4,5};
    
    char *char_pointer;
    int *int_pointer;
    
    char_pointer = (char*) int_array; /*typecast into the pointers data type*/
    int_pointer = (int*) char_array;
    
    for(i=0; i < 5; i++){
        printf("[integer pointer] points to %p, which is contains the
        char '%c'\n", int_pointer, *int_pointer);
    int_pointer = (int*) ((char*)int_pointer + 1);
    }
    
    for(i=0; i < 5; i++){
        printf("[char pointer] points to %p, which contains the int %d\n",
        char_pointer, *char_pointer);
    char_pointer = (char*) ((int*)char_pointer + 1);
    }
}

Can you list the line number for the errors ?
let me guess that they are 18 and 24

Fixed the errors and warnings.
NOTE: *int is NOT correct.

Tim S.

Code:

#include <stdio.h>

int main(){
    int i;

    char char_array[5] = {'a', 'b', 'c', 'd', 'e'};
    int int_array[5] = {1,2,3,4,5};

    char *char_pointer;
    int *int_pointer;

    char_pointer = (char*) int_array; /*typecast into the pointers data type*/
    int_pointer = (int*) char_array;

    for(i=0; i < 5; i++){
        printf("[integer pointer] points to %p, which is contains the char '%c'\n", int_pointer, *int_pointer);
        int_pointer = (int*) ((char*)int_pointer + 1);
    }

    for(i=0; i < 5; i++){
        printf("[char pointer] points to %p, which contains the int %d\n",
        char_pointer, *char_pointer);
        char_pointer = (char*) ((int*)char_pointer + 1);
    }

    return 0;
}

Originally Posted by prafiate

EDIT: I also dont understand the argument. isnt typecasting supposed to have this format: (typecast_data) variable ? so wouldnt it be just ((int*) char_pointer+1); ?

It appears to be doing a very poor job of teaching pointer math.

Tim S.