question about struct

[kalar]

In an example i found this
We have this struct:
[PHP]struct catalog {
char name[80];
char title[80];
char pub[80];
unsigned date;
unsigned char ed;
} cat[MAX];
[/PHP]

When we fill the cat[i].name i saw this
[PHP]
for(i=0 i<MAX; i++)
printf("Enter name..")
gets(cat[i].name);
if(!*cat[i].name) break;/* what this command do? I mean the !* ?[/PHP]
What is !* ?
Thanks in advance

[ahoyer]

When we fill the cat[i].name i saw this
[PHP]
for(i=0 i<MAX; i++)
printf("Enter name..")
gets(cat[i].name);
if(!*cat[i].name) break;/* what this command do? I mean the !* ?[/PHP]
What is !* ?

Hey im not too familiar with php or anything, but the ! is the not operator for conditional statements, as far as the * i think it might be dereferencing a member of the struct, just written unclearly.

so what i think theyre doing is just checking to see if the value that the user entered is null, and if it is then kick out of the loop.

thats my best guess anyways.

[weaknessforcats]

cat is an array of catalog struct variables.
cat[i] is one of those array elements.
cat[i].name is an array of char that is a member of one of those elements.

In C/C++, the name of an array is the address of element 0. Therefore:

cat[i].name is the address of element 0 of the char name array.

Therefore:

*cat[i].name is element 0 of the name array

Therefore:
!*cat[i].name is true if element 0 of the name array is false.

Element 0 of the name array is false if it is a '\0'.

Therefore:

!*cat[i].name is true of the name is a null string.

This could also have been written as:

if (cat[i].name[0] == '\0') break;

[kreagan]

In an example i found this

When we fill the cat[i].name i saw this
[PHP]
for(i=0 i<MAX; i++)
printf("Enter name..")
gets(cat[i].name);
if(!*cat[i].name) break;/* what this command do? I mean the !* ?[/PHP]
What is !* ?
Thanks in advance

The question should probably be: What is "!" and what is "*". It might make more sense to write that line:

[CODE=PHP]if ( !(*cat[i].name)) break;[/CODE]

The "!" means not and the * is most likely dereferencing a pointer. If there is no value in cat[i].name, then break.

This is only a guess. Some languages see primatives (int, char) as "TRUE". So if there is no value in cat[i].name, then *cat[i].name should return NULL. NULL is FALSE. Therefore !NULL = TRUE.

Hope this helps.

[kalar]

I also think about pointers but. When we have as example an array p[i] and we use a pointer pp for the array, we don't say *pp[i] but only *pp.Am i wrong?


edit:Maybe help you that the programm says that if we enter an empty line in the name breaks

[kreagan]

I also think about pointers but. When we have as example an array p[i] and we use a pointer pp for the array, we don't say *pp[i] but only *pp.Am i wrong?


edit:Maybe help you that the programm says that if we enter an empty line in the name breaks

Are you sure *pp.Am is how you use pointers for an array? With *pp[i].Am, you are stating ... get the value of AM of the ith element of pp.

pp.Am doesn't really make sense to me because there are multiple Ams. I could be mistaken though.

[JosAH]

Did anyone actually *read* post #3? WeaknessForCats actually explained what
that expression is all about. No need for guessing anymore.

kind regards,

Jos