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Editorial Team
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Asked: 2026-05-20T01:48:45+00:00

#include<stdio.h> #include<stdlib.h> #include<string.h> #include<stdbool.h> char** parser(char *message) { char a[9][256]; char* tmp =message; bool

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#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<stdbool.h>
char** parser(char *message)
{
    char a[9][256];
    char* tmp =message;
    bool inQuote=0;
    int counter=0;
    int counter2=0;
    while(*tmp!='\0')
    {
        switch(*tmp)
        {
            case ',': if(!inQuote)
                   {    
                    a[counter][counter2]='\0';
                    printf("A[%d]: %s\n",counter,a[counter]);
                    counter++;
                    counter2=0;
                    }

                  break;
            case '"':
                inQuote=!inQuote;
                break;
            default:
                a[counter][counter2]=*tmp;
                counter2++;
                break;
        }
    tmp++;

    }
        a[counter][counter2]='\0';
        printf("A[%d]: %s\n",counter,a[counter]);
        return(a);

}
int main()
{
    char **a = parser("N,8545,01/02/2011 09:15:01.815,\"RASTA OPTSTK 24FEB2011 1,150.00 CE\",S,8.80,250,0.00,0");
    return 0;
}

The error given is at Line 38 : return from incompatible pointer type and function returns address of local variable

EDIT Can somebody modify the code accordingly so that I can access (via pointer) the contents of ‘a’ from main();

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1 Answer

  1. Editorial Team

    The errors speak for themselves

    Line 38 : return from incompatible
    pointer type

    a has been defined as char a[9][256]. So, in the statement return(a);, the type of value being returned is char (*)[256] ( pointer to an array of 256 chars ) and not char ** ( as per the prototype of parser() )

    function returns address of local
    variable

    Well, a is a variable local to the function. You should not be returning address of local variables ( unless memory for it has been dynamically allocated or it is a static variable )

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