6
Why in the statement as argument in function main(): char* argv[] instead of char argv[] i can access the information. I know one is pointer. I can’t access when it’s not pointer.
#include <stdio.h>
int main (int argc, char* argv[])
{
printf("%s", argv[1]);
}
5
You are accessing a array of strings. The [] represents the array and the char *represents the string. This is necessary because the command line can pass several arguments, and all are strings.
In C there is no string with its own concept, it is usually represented by a array or the most common character pointer.
So it’s two different things, so it needs to be used this way.
In C it’s rare, but some do typedef string char *; to use string in place of char *.
If it were
int main(int argc, string argv[])
I put in the Github for future reference.
would you understand? It’s the same thing.
Can’t access without the pointer array of characters and not a array of strings, which is what is expected. Actually you can access, but it won’t be what you expect. You have to use the right type for what you need.
3
Because in C strings are character vectors, when you declare char* argv[], you are declaring a string array.
To access from index 1 to n-times, 0 and the name passed on the command line.
This is an example:
#include <stdio.h>
#include <stdlib.h>
int main()
{
char *nomes[] = {"joao", "maria", "pedro"};
for(int i=0; i<sizeof(nomes) / sizeof(char* ); i++) {
printf("%s\n", nomes[i]);
}
return 0;
}
1
Just to better understand, pointers are useful when a variable has to be accessed in different parts of a program. Your code can have multiple pointers "pointing" to the variable that contains the desired data.
Situations where pointers are useful:
Browser other questions tagged c pointer
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