Help with c errors

getting this error:

1>c:\users\b1021568\documents\visual

studio 2010\projects\tarefa42\tarefa

42\main.cpp(112): error C2664: 'cria_aluno' : cannot convert parameter 2 from 'c开发者_开发知识库onst char [7]' to 'char' 1> There is no context in which this conversion is possible

when trying to compile this:

int main(void)
{
    Aluno *a[5];

    a[0] = cria_aluno(1, "turma1", "Joao", 7.0, 8.4, 4.3);
    a[1] = cria_aluno(2, "turma2", "Maria", 3.2, 5.1, 10.0);
    a[2] = cria_aluno(3, "turma3", "Rafael", 8.1, 3.2, 4.5);
    a[3] = cria_aluno(4, "turma4", "Jose", 1.3, 7.7, 9.3);
    a[4] = cria_aluno(5, "turma5", "Lais", 4.5, 1.3, 9.9);

    ordena(5, a);
return 0;
}

thats my cria_aluno function:

Aluno *cria_aluno(int mat, char turma, char nome, float p1, float p2, float p3)
{
    Aluno *a;

    a = (Aluno*) malloc(sizeof(Aluno));
    if(a == NULL)
    {
        printf("Memoria insuficiente");
        return NULL;
    }
    a->mat = mat;
    a->turma = turma;
    strcpy(a->nome, nome);
    a->p1 = p1;
    a->p2 = p2;
    a->p3 = p3;
    return a;
}

Change it to

Aluno *cria_aluno(int mat, const char* turma, const char* nome, float p1, float p2, float p3)

"turma1", etc. are of type const char*, not char

Change

Aluno *cria_aluno(int mat, char turma, char nome, float p1, float p2, float p3)

to

Aluno *cria_aluno(int mat, const char* turma, const char* nome, float p1, 
                  float p2, float p3)
{
   Aluno *a = (Aluno*) malloc(sizeof(Aluno));
   if(a == NULL)
   {
      printf("Memoria insuficiente");
      return NULL;
   }
   a->mat = mat;
   a->turma = malloc(strlen(turma)+1);
   strcpy(a->turma, turma);
   a->nome = malloc(strlen(nome)+1);
   strcpy(a->nome, nome);
   a->p1 = p1;
   a->p2 = p2;
   a->p3 = p3;
   return a;
}

Your function expect as parameter 2 and 3 char type and not char pointer (char*, usually used as "string").

In you main function, you called cria_aluno with char* type (string) as parameter 2 and 3, that is the cause of your error.

First you need to decide what you wish to store in Aluno structure. Lets take turma as example:

If you wish to store a single character, you should use char as the type of turma in the structure and in the function. Also, in the function call, you should use a single character as parameter 2, for example: 'a'. To copy this character, you should use a simple copy: a->turma = turma;

If you wish to store a string, you should use char[x] (where x is the max string length + \0 at the end) as the type of turma in the structure. In the function, you should use char* (const char* will be better). In the function call, you can use a string (example: "example"). To copy this attribute, you should use strcpy.

Another way to store turma in your structure as string mode, is change the type to char* in the structure. Then, when needed, allocate the memory.

Good luck

Amir

In the function call

a[0] = cria_aluno(1, "turma1", "Joao", 7.0, 8.4, 4.3); 

"turma1" and "Joao" are string literals, which are arrays of char (const char in C++). The types of the two expressions are char [7] and char [5], respectively. These types are not compatible with char, which is what you've declared turma and nome to be in cria_aluno, hence the error.

In most circumstances, array expressions have their types implicitly converted from "N-element array of T" to "pointer to T". So what actually gets passed to cria_aluno are two expressions of type char *, not char. Thus, you need to change the declaration of cria_aluno to

Aluno *cria_aluno(int mat, const char *turma, const char *nome, float p1, float p2, float p3)

Why const char * instead of char *? This helps protect you from accidentally modifying the contents of what the pointer points to; attempting to modify the contents of a string literal leads to undefined behavior.