Home/ Questions/Error: assignment to expression with array type - how can i circumvent the error?
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Mia Noel
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Asked: 2022-05-17T12:49:57+00:00 In:

Error: assignment to expression with array type – how can i circumvent the error?

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I’m trying to run a new project. I do a couple of things like this:

#include <stdio.h>
 
 #define N 30
 
 typedef struct{
  char name[N];
  char surname[N];
  int age;
 } data;
 
 int main() {
  data s1;
  s1.name="Paolo";
  s1.surname = "Rossi";
  s1.age = 19;
  getchar();
  return 0;
 }
s1.name="Paolo" 
 s1.surname="Rossi"
data s1 = {"Paolo", "Rossi", 19};

But in my program, I am getting the warning:

"error: assignment to expression with array type error"

Can someone explain why the “error: assignment to expression with array type” issue happened? Where have I gone wrong? Thank you!

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2 Answers

  1. Best Answer
    Expert

    The cause: It shows the error “Not Assignable” because assigning a value to a string directly in C is not feasible.

    Solution:

    You should use strcpy() function to modify the value

    Do it in the following way:

    strcpy(s1.name , "Egzona"); 
printf( "Name : %s\n", s1.name);
    • 24
  2. typedef struct{
 char name[30];
 char surname[30];
 int age;
} data;

    data is a block of memory that can hold 60 characters plus 4 for the int. (See note). 

    [----------------------------,------------------------------,----]
 ^ this is name ^ this is surname ^ this is age

    This allocates memory to the stack. 

    data s1;

    Assignments are not just copies of numbers; sometimes they can be pointers.

    This is a failure 

    s1.name = "Paulo";

    Because s1.name is at the beginning of a 64-byte-long struct, and "Paulo" is 6 bytes (6 because of trailing 0 in C string strings), the compiler can assign a pointer for a string to a string.

    To copy “Paulo into struct at point name and “Rossi into struct at Point surname. 

    memcpy(s1.name, "Paulo", 6);
memcpy(s1.surname, "Rossi", 6);
s1.age = 1;

    What do you end up with? 

    [Paulo0----------------------,Rossi0-------------------------,0001]

    strcpy does the exact same thing, but it knows about \0 termination and does not require the length to be hardcoded.

    Alternativly, you can also define a structure that points at char arrays any length. 

    typedef struct {
 char *name;
 char *surname;
 int age;
} data;

    This will result in 

    [----,----,----]

    Now, you can fill the struct using pointers. 

    s1.name = "Paulo";
s1.surname = "Rossi";
s1.age = 1;

    This is what it looks like 

    [---4,--10,---1]

    Pointers are 4 and 10.

    Please note that the pointers and ints can have different sizes. For example, size 4 is 32bit.

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