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Home/ Questions/Q 6853289
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T01:29:40+00:00 2026-05-27T01:29:40+00:00

I know this is really easy and I’m looking over something but this is

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I know this is really easy and I’m looking over something but this is what I have…:

typedef struct
{
    char s1[81];
    char s2[81];
    char s3[81];
}Rec;

int main()
{   

   Rec *a[10];

   a[0] = (Rec*)new unsigned char(sizeof(Rec));
   a[0]->s1= "hello";
   printf("a[0] = %s\n",a[0]->s1);
   delete(a[0]);


   getchar();
   return 0;
}

Now, the line

a[0]->s1= “hello”;

is complaining about the expression must be a modifiable lvalue. I am pretty sure it’s how I’m casting it in my new operator line and has it needs to be a long value or something but I’m not sure of the code to do this… easy i know but yeah. Any help would be much appreciated

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  1. Editorial Team
    Editorial Team
    2026-05-27T01:29:41+00:00Added an answer on May 27, 2026 at 1:29 am

    You cannot assign to char arrays like that. Either use strcpy, or change your char arrays to std::string.

    strcpy(a[0]->s1, "hello");
    

    Why are you doing this:

    a[0] = (Rec*)new unsigned char(sizeof(Rec));
    

    instead of this:

    a[0] = new Rec;
    
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