strtod definition and type of passed pointer

Question

From definition double strtod ( const char * str, char ** endptr );

C reference sites provide an example for that mighty function:

char szOrbits[] = "365.24 29.53";
char * pEnd;
double d1, d2;
d1 = strtod (szOrbits,&pEnd);
d2 = strtod (pEnd,NULL);

If endptr should be of type char ** why is char *pEnd used here ? And not char **pEnd ?

`endptr` is of type `char**` so that the function could modify the `char*` it points to. — eq-, Sep 21 '12 at 19:27

score 6 · Accepted Answer · edited Sep 21 '12 at 20:40

6

The type of pEnd is char *. The type of &pEnd is char **.

C passes arguments to functions by value, so to modify a pointer you need to pass a pointer to the pointer.

You could define a char ** directly but you would need to initialize it to a char * as the idea is that strtod is modifying a char *.

Like:

char *q;
char **p = &q;

d1 = strtod (szOrbits, p);

Obviously the intermediate p pointer is not necessary and you can use &q directly.

edited Sep 21 '12 at 20:40

Jim Balter

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answered Sep 21 '12 at 19:29

ouah

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+1 nice explanation of a classic misunderstanding of pointers/ – R.. GitHub STOP HELPING ICE Sep 21 '12 at 22:32

score 2 · Answer 2 · answered Sep 21 '12 at 19:28

2

pEnd is a char*, as you've observed. The first example call to strtod() passes &pEnd as the second argument. &pEnd is the address of pEnd, and is therefore of type char**.

answered Sep 21 '12 at 19:28

chrisaycock

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strtod definition and type of passed pointer

2 Answers2