This error shows the error as not assignable due to the value associated with the string; or the value is not feasible in C programming language.
In order to solve this, you have to use a normal stcpy() function to modify the value.
However, there are other possible ways to fix this issue.
Causes of the error
Here are few reasons for error: assignment to expression with array type:
#1. Using unassignable array type
You are using an array type that is not assignable, or one that is trying to change the constant value in the program
If you are trying to use an unsignable array in the program, here’s what you will see:
#include
#define N 20
typedef struct{
char name[I];
char surname[I];
int age;
} data;
int main() {
data s1;
s1.name = “Ogechukwu”;
s1.surname = “Anthony”;
s1.age = 28;
getchar();
return 0;
}This is because s1.name is an array type that is unassignable. Therefore, the output contains an error.
#2. Error occurs because block of memory is below 60 chars
The error also appears if the block of memory is below 60 chars. The data should always be a block of memory that fits 60 chars.
In the example above, “data s1” is responsible for allocating the memory on the stack. Assignments then copy the number, and fails because the s1.name is the start of a struct 64 bytes long which can be detected by a compiler, also Ogechukwu is 6 bytes long char due to the trailing \0 in the C strings. So , assigning a pointer to a string into a string is impossible, and this expression error occurs.
Example:
memcspy(s1.name, “Ogechukwu”,6)
memcspy(s1.surname, “Anthony”,6)
s1.age=2;The output will be thus :
[Ogechukwu0—————–,Anthony0——————-,0002]
Error: assignment to expression with an array type.#3. Not filling structure with pointers
If a programmer does not define a structure which points towards char arrays of any length, an expression error occurs, however If the programmer uses a defined structure field, they also have to set pointers in it too. Example of when a program does not have a defined struct which points towards a char.
Errtypedef strcpy {
char *Kani;
char *Sundew;
int age;
} data;-,#4. There is a syntax error
If a Programmers uses a function in an incorrect order, or syntax.also if the programer typed in the wrong expression within the syntax.
example of a wrong syntax, where using “-” is invalid:
strcpy(s1-name , “Peace”);
printf( “Name – %s\n”, s1.name);#5. You directly assigned a value to a string
The C programming language does not allow users to assign the value to a string directly. So you cant directly assign value to a string, ifthey do there will receive an expression error.
Example of such:
struct (s1.name, “New_Name”);#6. Changing the constant value
As a programmer, if you are trying to change a constant value, you will encounter the expression error. an initialized string , such as “char sample[19]; is a constant pointer. Which means the user can change the value of the thing that the pointer is pointing to, but cannot change the value of the pointer itself. When the user tries to change the pointer, they are actually trying to change the string’s address, causing the error.
Example:
Program:include int main() {
char nam#e[19];
printf(“address of sample:%d address of a random string:%d”,name,”Ogechukwu”);
}The output gives the programmer two independent addresses. The first address is the address of s1.name while the second address is that of Ogechukwu. So a programmer cannot change s1.name address to Ogechukwu address because it is a constant.
How to fix “error: assignment to expression with array type” error?
The following will help you fix this error:
#1. Use an assignable array type
Programmers are advised to use the array type that is assignable by the arrays because an unassignable array will not be recognized by the program causing the error. To avoid this error, make sure the correct array is being used.
Check out this example:
#include
#define N 30
typedef struct{
char name[N];
char surname[N];
int age;
} data;
int main() {
data s1 = {“Ogechukwu”, “Anthony”, 20};
getchar();
return 0;
}#2. Use a block of memory that fits 60 chars
Use the struct function to define a struct pointing to char arrays of any lenght. This means the user does not have to worry about the length of the block of memory. Even if the length cis below 60 chars, the program will still run smoothly.
But if you are not using the struct function, take care of the char length and that the memory block should fit 60 chars and not below to avoid this error.
#3. Fill the structure with pointers
If the structure is not filled with pointers, you will experience this error. So ensure you fill the struct with pointers that point to char arrays.
Keep in mind that the ints and pointers can have different sizes. Example
Program:
typedef struct{
s1.name = “Ogechukwu”;
s1.surname = “Anthony”;
s1.age = 3;This example, the struct has been filled with pointers.
#4. Use the correct syntax
Double check and ensure you are using the right syntax before coding.if you are confused about the right syntax, then consult Google to know the right one.
Example
.strcpy(s1.name , “Ogechukwu”);
printf( “Name : %s\n”, s1.name);#5. Use Strcpy() function to modify
In order to modify the pointer or overall program, it is advisable to use the strcpy() function. This function will modify the value as well because directly assigning the value to a string is impossible in a C programming language. usin g this function helps the user get rid of assignable errors. Example
Strcpy(s1.name, “New_Name”);
#6. Copy an array using Memcpy
In order to eliminate this error,a programmer can use the memcpy function to copy an array into a string and prevent errors.
Example:
float transform[7] = {0,0,0,0,0,0,0};
struct Image *ex = calloc(1, sizeof *ex);
ex->name = “tests”;
memcpy(ex->transform, transform, sizeof ex->transform);#7. Copy into the array using Strcpy()
Also you can copy into the array using strcpy() because of the modifiable lvalue. A modifiable lvalue is an lvalue that has no array type. So programmers have to use strcpy() to copy data into an array.
Follow these guide properly and get rid of “error: assignment to expression with array type”.
