why temporary char** argument is ilegal?

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7
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I have a function f(char **p) and I wanted to call it in the simplest way I could.

I tried char ** p = "a", "b"; f(p); and got:

scalar object requires one element in initializer

so I changed into char * p[2] = "a", "b"; f(p); and that went fine [would have been fine with just char * p].

My question is, why can't I create an array of pointers on the fly like this:

f("a","b");

edited 7 mins ago

asked 16 mins ago

CIsForCookies

6,50411441

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up vote
7
down vote

favorite

I have a function f(char **p) and I wanted to call it in the simplest way I could.

I tried char ** p = "a", "b"; f(p); and got:

scalar object requires one element in initializer

so I changed into char * p[2] = "a", "b"; f(p); and that went fine [would have been fine with just char * p].

My question is, why can't I create an array of pointers on the fly like this:

f("a","b");

edited 7 mins ago

asked 16 mins ago

CIsForCookies

6,50411441

add a commentÂ |Â

up vote
7
down vote

favorite

I have a function f(char **p) and I wanted to call it in the simplest way I could.

I tried char ** p = "a", "b"; f(p); and got:

scalar object requires one element in initializer

so I changed into char * p[2] = "a", "b"; f(p); and that went fine [would have been fine with just char * p].

My question is, why can't I create an array of pointers on the fly like this:

f("a","b");

edited 7 mins ago

asked 16 mins ago

CIsForCookies

6,50411441

I have a function f(char **p) and I wanted to call it in the simplest way I could.

I tried char ** p = "a", "b"; f(p); and got:

scalar object requires one element in initializer

so I changed into char * p[2] = "a", "b"; f(p); and that went fine [would have been fine with just char * p].

My question is, why can't I create an array of pointers on the fly like this:

f("a","b");

edited 7 mins ago

asked 16 mins ago

CIsForCookies

6,50411441

edited 7 mins ago

asked 16 mins ago

CIsForCookies

6,50411441

edited 7 mins ago

asked 16 mins ago

CIsForCookies

6,50411441

asked 16 mins ago

CIsForCookies

6,50411441

asked 16 mins ago

CIsForCookies

6,50411441

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
6
down vote

This gives a warning:

char ** p = "a", "b";

Because p is not an array.

This is also not legal:

f({"a", "b"));

Since curly braces by themselves are not allowed in an expression (but can be used as an initializer).

It is possible to create an array on the fly like that using a compound literal:

f((char *)"a", "b");

You could also use a compound literal to initialize a temporary:

char ** p = (char *)"a", "b";

Unlike the first statement, this is valid because an array of type char *[2] will decay to a char ** and can be used to initialize a variable of this type.

See section 6.5.2.5 of the C standard for more details on compound literals.

edited 3 mins ago

answered 13 mins ago

dbush

86.6k1091124

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up vote
3
down vote

This declaration char ** p = "a", "b"; triggers a warning because C does not know how to interpret the content of braces:

Declaration type char** suggests it's a single pointer-to-pointer-to-char

Content of specifies two items.

You need to tell C that you are initializing a pointer to pointer with a pointer to the initial element of an anonymous array by specifying a type in front of the braces:

char ** p = (char*)"a", "b"

This construct is called a "compound literal". It could be used in a declaration, but it also works as an expression.

Note: if you are planning to pass an array like that to a function, you need to provide a way to determine the size of the array. One approach is to pass NULL to "terminate" the array, like this:

void f(char **p) 
 int i = 0;
 while (*p) 
 printf("%d:%sn", i++, *p++);
 

int main(void) 
 f((char*)"a", "b", NULL);
 return 0;

Demo.

answered 10 mins ago

dasblinkenlight

603k547591174

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up vote
0
down vote

C99 and later added compound literals for the exact purpose: creating array and structs on the fly

Example:

struct foo int a; char b[2]; structure;
structure = ((struct foo) 1, 'a', 0);
int y = (int ) 1, 2, 3;
int z = (int [3]) 1;

Apart from std C99 and later, GCC provide this feature as an extension too.

In you case this will work

f((char *)"a","b");

char *)"a","b" is a compound literal which creates an array of 2 pointers to char.

edited 11 secs ago

answered 5 mins ago

haccks

84.7k20125216

add a commentÂ |Â

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
6
down vote

This gives a warning:

char ** p = "a", "b";

Because p is not an array.

This is also not legal:

f({"a", "b"));

Since curly braces by themselves are not allowed in an expression (but can be used as an initializer).

It is possible to create an array on the fly like that using a compound literal:

f((char *)"a", "b");

You could also use a compound literal to initialize a temporary:

char ** p = (char *)"a", "b";

Unlike the first statement, this is valid because an array of type char *[2] will decay to a char ** and can be used to initialize a variable of this type.

See section 6.5.2.5 of the C standard for more details on compound literals.

edited 3 mins ago

answered 13 mins ago

dbush

86.6k1091124

add a commentÂ |Â

up vote
6
down vote

This gives a warning:

char ** p = "a", "b";

Because p is not an array.

This is also not legal:

f({"a", "b"));

Since curly braces by themselves are not allowed in an expression (but can be used as an initializer).

It is possible to create an array on the fly like that using a compound literal:

f((char *)"a", "b");

You could also use a compound literal to initialize a temporary:

char ** p = (char *)"a", "b";

Unlike the first statement, this is valid because an array of type char *[2] will decay to a char ** and can be used to initialize a variable of this type.

See section 6.5.2.5 of the C standard for more details on compound literals.

edited 3 mins ago

answered 13 mins ago

dbush

86.6k1091124

add a commentÂ |Â

up vote
6
down vote

This gives a warning:

char ** p = "a", "b";

Because p is not an array.

This is also not legal:

f({"a", "b"));

Since curly braces by themselves are not allowed in an expression (but can be used as an initializer).

It is possible to create an array on the fly like that using a compound literal:

f((char *)"a", "b");

You could also use a compound literal to initialize a temporary:

char ** p = (char *)"a", "b";

Unlike the first statement, this is valid because an array of type char *[2] will decay to a char ** and can be used to initialize a variable of this type.

See section 6.5.2.5 of the C standard for more details on compound literals.

edited 3 mins ago

answered 13 mins ago

dbush

86.6k1091124

This gives a warning:

char ** p = "a", "b";

Because p is not an array.

This is also not legal:

f({"a", "b"));

Since curly braces by themselves are not allowed in an expression (but can be used as an initializer).

It is possible to create an array on the fly like that using a compound literal:

f((char *)"a", "b");

You could also use a compound literal to initialize a temporary:

char ** p = (char *)"a", "b";

Unlike the first statement, this is valid because an array of type char *[2] will decay to a char ** and can be used to initialize a variable of this type.

See section 6.5.2.5 of the C standard for more details on compound literals.

edited 3 mins ago

answered 13 mins ago

dbush

86.6k1091124

edited 3 mins ago

answered 13 mins ago

dbush

86.6k1091124

answered 13 mins ago

dbush

86.6k1091124

answered 13 mins ago

dbush

86.6k1091124

add a commentÂ |Â

up vote
3
down vote

This declaration char ** p = "a", "b"; triggers a warning because C does not know how to interpret the content of braces:

Declaration type char** suggests it's a single pointer-to-pointer-to-char

Content of specifies two items.

You need to tell C that you are initializing a pointer to pointer with a pointer to the initial element of an anonymous array by specifying a type in front of the braces:

char ** p = (char*)"a", "b"

This construct is called a "compound literal". It could be used in a declaration, but it also works as an expression.

void f(char **p) 
 int i = 0;
 while (*p) 
 printf("%d:%sn", i++, *p++);
 

int main(void) 
 f((char*)"a", "b", NULL);
 return 0;

Demo.

answered 10 mins ago

dasblinkenlight

603k547591174

add a commentÂ |Â

up vote
3
down vote

This declaration char ** p = "a", "b"; triggers a warning because C does not know how to interpret the content of braces:

Declaration type char** suggests it's a single pointer-to-pointer-to-char

Content of specifies two items.

You need to tell C that you are initializing a pointer to pointer with a pointer to the initial element of an anonymous array by specifying a type in front of the braces:

char ** p = (char*)"a", "b"

This construct is called a "compound literal". It could be used in a declaration, but it also works as an expression.

void f(char **p) 
 int i = 0;
 while (*p) 
 printf("%d:%sn", i++, *p++);
 

int main(void) 
 f((char*)"a", "b", NULL);
 return 0;

Demo.

answered 10 mins ago

dasblinkenlight

603k547591174

add a commentÂ |Â

up vote
3
down vote

This declaration char ** p = "a", "b"; triggers a warning because C does not know how to interpret the content of braces:

Declaration type char** suggests it's a single pointer-to-pointer-to-char

Content of specifies two items.

You need to tell C that you are initializing a pointer to pointer with a pointer to the initial element of an anonymous array by specifying a type in front of the braces:

char ** p = (char*)"a", "b"

This construct is called a "compound literal". It could be used in a declaration, but it also works as an expression.

void f(char **p) 
 int i = 0;
 while (*p) 
 printf("%d:%sn", i++, *p++);
 

int main(void) 
 f((char*)"a", "b", NULL);
 return 0;

Demo.

answered 10 mins ago

dasblinkenlight

603k547591174

This declaration char ** p = "a", "b"; triggers a warning because C does not know how to interpret the content of braces:

Declaration type char** suggests it's a single pointer-to-pointer-to-char

Content of specifies two items.

You need to tell C that you are initializing a pointer to pointer with a pointer to the initial element of an anonymous array by specifying a type in front of the braces:

char ** p = (char*)"a", "b"

This construct is called a "compound literal". It could be used in a declaration, but it also works as an expression.

void f(char **p) 
 int i = 0;
 while (*p) 
 printf("%d:%sn", i++, *p++);
 

int main(void) 
 f((char*)"a", "b", NULL);
 return 0;

Demo.

answered 10 mins ago

dasblinkenlight

603k547591174

answered 10 mins ago

dasblinkenlight

603k547591174

answered 10 mins ago

dasblinkenlight

603k547591174

answered 10 mins ago

dasblinkenlight

603k547591174

add a commentÂ |Â

up vote
0
down vote

C99 and later added compound literals for the exact purpose: creating array and structs on the fly

Example:

struct foo int a; char b[2]; structure;
structure = ((struct foo) 1, 'a', 0);
int y = (int ) 1, 2, 3;
int z = (int [3]) 1;

Apart from std C99 and later, GCC provide this feature as an extension too.

In you case this will work

f((char *)"a","b");

char *)"a","b" is a compound literal which creates an array of 2 pointers to char.

edited 11 secs ago

answered 5 mins ago

haccks

84.7k20125216

add a commentÂ |Â

up vote
0
down vote

C99 and later added compound literals for the exact purpose: creating array and structs on the fly

Example:

struct foo int a; char b[2]; structure;
structure = ((struct foo) 1, 'a', 0);
int y = (int ) 1, 2, 3;
int z = (int [3]) 1;

Apart from std C99 and later, GCC provide this feature as an extension too.

In you case this will work

f((char *)"a","b");

char *)"a","b" is a compound literal which creates an array of 2 pointers to char.

edited 11 secs ago

answered 5 mins ago

haccks

84.7k20125216

add a commentÂ |Â

up vote
0
down vote

C99 and later added compound literals for the exact purpose: creating array and structs on the fly

Example:

struct foo int a; char b[2]; structure;
structure = ((struct foo) 1, 'a', 0);
int y = (int ) 1, 2, 3;
int z = (int [3]) 1;

Apart from std C99 and later, GCC provide this feature as an extension too.

In you case this will work

f((char *)"a","b");

char *)"a","b" is a compound literal which creates an array of 2 pointers to char.

edited 11 secs ago

answered 5 mins ago

haccks

84.7k20125216

C99 and later added compound literals for the exact purpose: creating array and structs on the fly

Example:

struct foo int a; char b[2]; structure;
structure = ((struct foo) 1, 'a', 0);
int y = (int ) 1, 2, 3;
int z = (int [3]) 1;

Apart from std C99 and later, GCC provide this feature as an extension too.

In you case this will work

f((char *)"a","b");

char *)"a","b" is a compound literal which creates an array of 2 pointers to char.

edited 11 secs ago

answered 5 mins ago

haccks

84.7k20125216

edited 11 secs ago

answered 5 mins ago

haccks

84.7k20125216

answered 5 mins ago

haccks

84.7k20125216

answered 5 mins ago

haccks

84.7k20125216

add a commentÂ |Â

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