C language assignment to expression with array type [duplicate]

Question

I guys I have this struct

typedef struct objeto livro;
struct objeto
{
    char titulo[100], autor[100];
    int cota;
};

And this code:

int main()
{
FILE *f;
int tam=5, i=0;
livro c[5];
char a={"a"};

f == fopen("bib.dat", "wb");

if(f == NULL)
{
    printf("erro ao abrir ficheiro");
}

for(i=0; i<tam; i++)
{
    (*c).titulo=a;
    (*c).autor=a;
    (*c).cota=i;
    fwrite(c, sizeof(c), 1,f);
}

return 0;
}

but it says error:assignment to expression with array type on:

(*c).titulo=a;
(*c).autor=a;

I have tried everything I've seen here in the posts my i cant make it work

a is a single char. titulo is an array of chars. Therefore the error says you can’t assign them, which is correct. You probably want to use some variant of strncpy() — Sami Kuhmonen
– Sami Kuhmonen, Commented May 29, 2018 at 15:41
I think you should learn C language. That's the way the language goes: you cannot assign to an array, full stop. You can assign to pointers, or copy arrays with memcopy, but never try to assign to an array. BTW, char a={"a"}; is a horror: a is a single character while "a" is a string literal that is the const array {'a', '\0'}. Please use char a = 'a'; for a single char or char a[] = "a" for a C string. — Serge Ballesta
– Serge Ballesta, Commented May 29, 2018 at 15:43
The char a={"a"}; is probably giving you a compiler warning, if not an error. If you want a to be an initialized array of characters containing a string, declare it as char a[] = "a";. And as pointed out by Sami Kuhmonen, you cannot use an array as the left side of an assignment. — Ian Abbott
– Ian Abbott, Commented May 29, 2018 at 15:49
Did you want c[i] instead of (*c) (which is the same as c[0])? — Toby Speight
– Toby Speight, Commented May 29, 2018 at 16:00

Achal · Accepted Answer · 2018-05-29 16:02:16Z

2

Firstly, char a = {"a"}; is wrong as a is a single char variable but double quotation "a" means thats a string literal. To overcome this error first declare a is a char array like

char a[10] = "a";

And then use strcpy() or strncpy(), as (*c).titulo=a won't work because a is now declared as array & assigning one char array to another char array directly like (*c).titulo=a will try to change base address of titulo which is not possible.

Replace below statement

(*c).titulo=a;
(*c).autor=a;

with

strcpy((*c).titulo,a);
strcpy((*c).autor,a);

Secondly, the statement

f == fopen("bib.dat", "wb");

is wrong, probably you want to use = instead of ==

f = fopen("bib.dat", "wb");

edited May 29, 2018 at 16:02

answered May 29, 2018 at 15:50

Achal

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2 Comments

Sourav Ghosh Over a year ago

{} means array of characters..you may want to rephrase. say, {1,2,3}

Achal Over a year ago

Thanks. I modified @SouravGhosh

Leonard · Accepted Answer · 2018-05-29 15:47:58Z

0

Use the library function strncpy()for strings.

answered May 29, 2018 at 15:47

Leonard

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