Segfault while accessing dynamicly allocated C struct array

-1

I have the following struct that is defined in the following way

typedef struct _abcd {

    int a;

    unsigned long b;  

    void (*c)(int);

    int d;

} abcd_t, *abcd;

Now i have the following code

static abcd foo



int set_size(int size){

   foo = malloc(sizeof(abcd) * size);

}

This code for some reason gives me segfault when accessing some of the properties of array members.
But i have noticed that if i change the malloc line to the following - it fixes the issue

foo = malloc(sizeof(foo[0]) * size);

I find it strange as obviously sizeof(foo[0]) = sizeof(abcd)
So what is exactly the difference here?

Thanks

edited Nov 15 '18 at 0:56

asked Nov 15 '18 at 0:50

Dor Dali

235

add a comment |

-1

I have the following struct that is defined in the following way

typedef struct _abcd {

    int a;

    unsigned long b;  

    void (*c)(int);

    int d;

} abcd_t, *abcd;

Now i have the following code

static abcd foo



int set_size(int size){

   foo = malloc(sizeof(abcd) * size);

}

This code for some reason gives me segfault when accessing some of the properties of array members.
But i have noticed that if i change the malloc line to the following - it fixes the issue

foo = malloc(sizeof(foo[0]) * size);

I find it strange as obviously sizeof(foo[0]) = sizeof(abcd)
So what is exactly the difference here?

Thanks

edited Nov 15 '18 at 0:56

asked Nov 15 '18 at 0:50

Dor Dali

235

add a comment |

-1

I have the following struct that is defined in the following way

typedef struct _abcd {

    int a;

    unsigned long b;  

    void (*c)(int);

    int d;

} abcd_t, *abcd;

Now i have the following code

static abcd foo



int set_size(int size){

   foo = malloc(sizeof(abcd) * size);

}

This code for some reason gives me segfault when accessing some of the properties of array members.
But i have noticed that if i change the malloc line to the following - it fixes the issue

foo = malloc(sizeof(foo[0]) * size);

I find it strange as obviously sizeof(foo[0]) = sizeof(abcd)
So what is exactly the difference here?

Thanks

edited Nov 15 '18 at 0:56

asked Nov 15 '18 at 0:50

Dor Dali

235

I have the following struct that is defined in the following way

typedef struct _abcd {

    int a;

    unsigned long b;  

    void (*c)(int);

    int d;

} abcd_t, *abcd;

Now i have the following code

static abcd foo



int set_size(int size){

   foo = malloc(sizeof(abcd) * size);

}

This code for some reason gives me segfault when accessing some of the properties of array members.
But i have noticed that if i change the malloc line to the following - it fixes the issue

foo = malloc(sizeof(foo[0]) * size);

I find it strange as obviously sizeof(foo[0]) = sizeof(abcd)
So what is exactly the difference here?

Thanks

c malloc dynamic-arrays

edited Nov 15 '18 at 0:56

asked Nov 15 '18 at 0:50

Dor Dali

235

edited Nov 15 '18 at 0:56

asked Nov 15 '18 at 0:50

Dor Dali

235

edited Nov 15 '18 at 0:56

asked Nov 15 '18 at 0:50

Dor Dali

235

asked Nov 15 '18 at 0:50

Dor Dali

235

asked Nov 15 '18 at 0:50

Dor Dali

235

add a comment |

2 Answers
2

active

oldest

votes

obviously sizeof(foo[0]) = sizeof(abcd)

It is not the same since you typedefed abcd to be a *pointer* to struct _abcd.

Use

foo = malloc(sizeof(*foo) * size);

to have robust code even if the type of foo should change at some point.

Your

foo = malloc(sizeof(foo[0]) * size);

is essentially the same since foo[0] is just syntactic sugar for *(foo + 0) which becomes *foo.

answered Nov 15 '18 at 0:55

Swordfish

9,71011436

add a comment |

The bug is you're allocating sizeof(abcd) and that's a pointer, not a struct. You want sizeof(abcd_t) or sizeof(*abcd).

answered Nov 15 '18 at 0:56

John Zwinck

153k17177294

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

obviously sizeof(foo[0]) = sizeof(abcd)

It is not the same since you typedefed abcd to be a *pointer* to struct _abcd.

Use

foo = malloc(sizeof(*foo) * size);

to have robust code even if the type of foo should change at some point.

Your

foo = malloc(sizeof(foo[0]) * size);

is essentially the same since foo[0] is just syntactic sugar for *(foo + 0) which becomes *foo.

answered Nov 15 '18 at 0:55

Swordfish

9,71011436

add a comment |

obviously sizeof(foo[0]) = sizeof(abcd)

It is not the same since you typedefed abcd to be a *pointer* to struct _abcd.

Use

foo = malloc(sizeof(*foo) * size);

to have robust code even if the type of foo should change at some point.

Your

foo = malloc(sizeof(foo[0]) * size);

is essentially the same since foo[0] is just syntactic sugar for *(foo + 0) which becomes *foo.

answered Nov 15 '18 at 0:55

Swordfish

9,71011436

add a comment |

obviously sizeof(foo[0]) = sizeof(abcd)

It is not the same since you typedefed abcd to be a *pointer* to struct _abcd.

Use

foo = malloc(sizeof(*foo) * size);

to have robust code even if the type of foo should change at some point.

Your

foo = malloc(sizeof(foo[0]) * size);

is essentially the same since foo[0] is just syntactic sugar for *(foo + 0) which becomes *foo.

answered Nov 15 '18 at 0:55

Swordfish

9,71011436

obviously sizeof(foo[0]) = sizeof(abcd)

It is not the same since you typedefed abcd to be a *pointer* to struct _abcd.

Use

foo = malloc(sizeof(*foo) * size);

to have robust code even if the type of foo should change at some point.

Your

foo = malloc(sizeof(foo[0]) * size);

is essentially the same since foo[0] is just syntactic sugar for *(foo + 0) which becomes *foo.

answered Nov 15 '18 at 0:55

Swordfish

9,71011436

answered Nov 15 '18 at 0:55

Swordfish

9,71011436

answered Nov 15 '18 at 0:55

Swordfish

9,71011436

answered Nov 15 '18 at 0:55

Swordfish

9,71011436

add a comment |

The bug is you're allocating sizeof(abcd) and that's a pointer, not a struct. You want sizeof(abcd_t) or sizeof(*abcd).

answered Nov 15 '18 at 0:56

John Zwinck

153k17177294

add a comment |

The bug is you're allocating sizeof(abcd) and that's a pointer, not a struct. You want sizeof(abcd_t) or sizeof(*abcd).

answered Nov 15 '18 at 0:56

John Zwinck

153k17177294

add a comment |

The bug is you're allocating sizeof(abcd) and that's a pointer, not a struct. You want sizeof(abcd_t) or sizeof(*abcd).

answered Nov 15 '18 at 0:56

John Zwinck

153k17177294

The bug is you're allocating sizeof(abcd) and that's a pointer, not a struct. You want sizeof(abcd_t) or sizeof(*abcd).

answered Nov 15 '18 at 0:56

John Zwinck

153k17177294

answered Nov 15 '18 at 0:56

John Zwinck

153k17177294

answered Nov 15 '18 at 0:56

John Zwinck

153k17177294

answered Nov 15 '18 at 0:56

John Zwinck

153k17177294

add a comment |

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