how to use a struct pointer to get a struct pointer array's member
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typedef struct {
int s1;
int s2;
int s3
}strTst;strTst spst1 = {3,5,7};
strTst spst2 = {1,2,4};
strTst *Sp1[2] =
{
&spst1,
&spst2
};strTst *pt = &Sp1;
I wonder how can i use pt to get a value in spst2, I checked, (*pt) value = &spst1, *(pt+1) = &spst2. but the compiler not let me to use (*pt)->s1, or ( (strTst *)(*pt) )->s1
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typedef struct {
int s1;
int s2;
int s3
}strTst;strTst spst1 = {3,5,7};
strTst spst2 = {1,2,4};
strTst *Sp1[2] =
{
&spst1,
&spst2
};strTst *pt = &Sp1;
I wonder how can i use pt to get a value in spst2, I checked, (*pt) value = &spst1, *(pt+1) = &spst2. but the compiler not let me to use (*pt)->s1, or ( (strTst *)(*pt) )->s1
The last line won't compile:
Sp1is an array of 2 pointers tostrTstinstances, not astrTst*. To accesss1inspst2, you needstrTst *pt = Sp1[1];
pt->s1 = 0;Robust Services Core | Software Techniques for Lemmings | Articles
The fox knows many things, but the hedgehog knows one big thing. -
The last line won't compile:
Sp1is an array of 2 pointers tostrTstinstances, not astrTst*. To accesss1inspst2, you needstrTst *pt = Sp1[1];
pt->s1 = 0;Robust Services Core | Software Techniques for Lemmings | Articles
The fox knows many things, but the hedgehog knows one big thing. -
The last line won't compile:
Sp1is an array of 2 pointers tostrTstinstances, not astrTst*. To accesss1inspst2, you needstrTst *pt = Sp1[1];
pt->s1 = 0;Robust Services Core | Software Techniques for Lemmings | Articles
The fox knows many things, but the hedgehog knows one big thing.Can I cast the strTst* pt to an array of 2 pointers to strTst instance? with
strTst *pt = &Sp1;
although, there is a warning compiler issued. I get the address of the Sp1 array. I think since i get the address of a data object,then i can operate on the data object.
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Can I cast the strTst* pt to an array of 2 pointers to strTst instance? with
strTst *pt = &Sp1;
although, there is a warning compiler issued. I get the address of the Sp1 array. I think since i get the address of a data object,then i can operate on the data object.
A cast should only be used when necessary, which isn't the case here. EDIT: I'm compiling under C++, so maybe it's an error there, and only a warning in C.
Robust Services Core | Software Techniques for Lemmings | Articles
The fox knows many things, but the hedgehog knows one big thing. -
typedef struct {
int s1;
int s2;
int s3
}strTst;strTst spst1 = {3,5,7};
strTst spst2 = {1,2,4};
strTst *Sp1[2] =
{
&spst1,
&spst2
};strTst *pt = &Sp1;
I wonder how can i use pt to get a value in spst2, I checked, (*pt) value = &spst1, *(pt+1) = &spst2. but the compiler not let me to use (*pt)->s1, or ( (strTst *)(*pt) )->s1
How about this:
typedef struct {
int s1;
int s2;
int s3;
}strTst;strTst spst1 = { 3,5,7 };
strTst spst2 = { 1,2,4 };
strTst* Sp1[2] =
{
&spst1,
&spst2
};strTst** pt = Sp1;
void f () {
pt[1]->s1 = 0;
}Mircea
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typedef struct {
int s1;
int s2;
int s3
}strTst;strTst spst1 = {3,5,7};
strTst spst2 = {1,2,4};
strTst *Sp1[2] =
{
&spst1,
&spst2
};strTst *pt = &Sp1;
I wonder how can i use pt to get a value in spst2, I checked, (*pt) value = &spst1, *(pt+1) = &spst2. but the compiler not let me to use (*pt)->s1, or ( (strTst *)(*pt) )->s1
You have declared
Sp1to be an array of pointers, each of which points to astrTststructure. So to get one of the pointers you just need normal array addressing, e.g.Sp1[1]returns the second element of the array. So all you need is:strTst *pt = Sp1[1]; // pt now equals &spst2
int value = pt->s2; // get the value of spst2.s2 -
typedef struct {
int s1;
int s2;
int s3
}strTst;strTst spst1 = {3,5,7};
strTst spst2 = {1,2,4};
strTst *Sp1[2] =
{
&spst1,
&spst2
};strTst *pt = &Sp1;
I wonder how can i use pt to get a value in spst2, I checked, (*pt) value = &spst1, *(pt+1) = &spst2. but the compiler not let me to use (*pt)->s1, or ( (strTst *)(*pt) )->s1
strTst *pt = Sp1[0];
pt->s1; // is spst1.s1
pt->s2; // is spst1.s2
pt->s3; // is spst1.s3pt = Sp1[1];
pt->s1; // is spst2.s1
pt->s2; // is spst2.s2
pt->s3; // is spst2.s3// alternative
strTst **ppt = Sp1;
ppt[0]->s1; // is spst1.s1
ppt[0]->s2; // is spst1.s2
ppt[0]->s3; // is spst1.s3ppt[1]->s1; // is spst2.s1
ppt[1]->s2; // is spst2.s2
ppt[1]->s3; // is spst2.s3
