problem with the use of C pointers and step by reference

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problem with the use of C pointers and step by reference

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#1 by (2 votes)
#2 by (1 votes)
#3 by (1 votes)
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I do not understand why you do not put the asterisk in int end = strlen(str) - 1; , would not it be int end = strlen(*str) - 1; ?

void trim(char *str) // Remove whitespace from a string and properly null-
terminate it.
{
int i;
int begin = 0;
int end = strlen(str) - 1;
while (isspace(str[begin])) begin++;
while ((end >= begin) && isspace(str[end])) end--;
for (i = begin; i <= end; i++) str[i - begin] = str[i];
str[i - begin] = 'void trim(char *str) // Remove whitespace from a string and properly null-
terminate it.
{
int i;
int begin = 0;
int end = strlen(str) - 1;
while (isspace(str[begin])) begin++;
while ((end >= begin) && isspace(str[end])) end--;
for (i = begin; i <= end; i++) str[i - begin] = str[i];
str[i - begin] = '%pre%';
}
';
}

I understand that putting pointers per reference step you need a pointer.

c

asked by thc 08.03.2018 в 01:10

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4 answers

Generate getters and setters in ECLIPSE how can I resolve an error ... @ 15db9742

score 2 · Answer 1

In C there is no string type like in other languages. There is no data type to store text, chars arrays are used, which work just like the other arrays. We see curious things like for example that the symbol & is not used in the scanf. It is not necessary because it is an array, and we already know that writing the name of the array is equivalent to putting & name [0].

Therefore, if you look at the declaration of the function you are examining, you are receiving a pointer to a memory address, precisely to the first position of the string, that is to say that "str" points to the first character, if you print it as

printf("%p", str);

You will see its address, however, if you print it as

printf("%s", str);

You will see its contents until you find the NULL or end of string character '\ 0'

Greetings.

EDIT: If you would use the asterisk, you would get the first character of the string, that is, if you try

printf("%c", *str);

You would see the first character

score 1 · Answer 2

The pointers are one of the great potentials of C and C ++ because they come from C because I inherit them, but it is a characteristic of C. You have a confusion in the way pointers work, the pointers are a type of data which only a memory address can be saved, that is, pointers (pointers) to a memory address, a memory address that has already been assigned to another variable or dynamically ( Memory Management ), but it must be a memory address that exists and has already been assigned . When you define a pointer it always has the characteristic asterisk before the name, and to use it you can do it in different ways, besides with the pointers you can use something called Arithmetic of pointers , with an extremely basic example that may help put you in context:

int numeroINT = 100;
double numeroD = 50.025;

const char *texto = "En C no existe el objeto de alto nivel llamado String";

int *ptrInt = &numeroINT;
double *ptrD = &numeroD;

As you can see, I declare two variables, a int and the other type double , and a constant pointer to char which points to a text string, which can not be changed because it is a constant pointer .

Two pointers are declared, one to type int and another to type double , and these are passed the memory address of two variables of the same type, so those pointers are pointing to the memory address of those two variables.

Now with these two pointers you can modify the two variables, because the pointers have their memory address already assigned, following the previous example:

void imprimir_texto(const char *texto)
{
    printf("%s\n", texto);
}

void aumentar_numero(int *numero)
{
    ++(*numero);
}

imprimir_texto(texto);
//Estas dos llamadas son equivalentes
aumentar_numero(ptrInt);
aumentar_numero(&numeroINT);

Take a look at these two functions, one receives a constant pointer to char and prints it, and the other receives a pointer or memory address to a data type int , and its value increases it by 1 .

When a pointer is used, it is done in two ways, one is to use it without the asterisk, this means that it is being called to the memory address, but if you put the asterisk, you are calling it at < strong> value saved in that direction, are two very different pods.

What are the advantages of using pointers, it is a question I asked when I saw them for the first time, suppose that you have an incredibly large and heavy structure and it requires that you go through many functions, the values passed to functions are passed by value, that is, that a copy of the original object is made and the copy is passed, and if your structure is very large, that copy will take its time, considerably lowering the performance, a simple solution to this is to pass the memory address (pointer) of the structure , with this the original is passed instead of making a copy, adding this you can protect the original structure against copy by adding the reserved word const in front of the definition, in the same way that I did with the text.

score 1 · Answer 3

Would not it be int end = strlen(*str) - 1; ?

str is a pointer to char
*str is a char

The strlen function needs to receive a pointer to char to be able to traverse the string to the end and know its length.

So no, you should not use the asterisk in this case.

score 0 · Answer 4

The asterisk is only used when declaring or modifying a pointer.

Example:

char *cadena = "texto original"; //asigno
char **direccionDe = &cadena;
(*direccionDe) = "nuevo texto"; //modifico
printf("%s", cadena); //nuevo texto

In your example, you are only obtaining the size of said chain and subtracting one from it.