Why this is wrong?

[linq936]

Hi,
I have this code,

unsigned int input_0[8];

*input_0++ = 100;
*input_0++ = 200;
*input_0++ = 300;
*input_0++ = 400;
*input_0++ = 1;
*input_0++ = 2;
*input_0++ = 3;
*input_0 = 4;

The compiler is gcc ported to a specific embedded processor. The
compiler complains that I can not use increment operation like that.

Do you know why?

 

[Rafael Almeida]

On 13 Jun 2006 17:13:46 -0700

Hi,
I have this code,

unsigned int input_0[8];

*input_0++ = 100;
*input_0++ = 200;
*input_0++ = 300;
*input_0++ = 400;
*input_0++ = 1;
*input_0++ = 2;
*input_0++ = 3;
*input_0 = 4;

The compiler is gcc ported to a specific embedded processor. The
compiler complains that I can not use increment operation like that.

Do you know why?

Because the C standard doesn't say it should be allowed.

 

[Richard Tobin]

unsigned int input_0[8];

*input_0++ = 100;

input_0 is an array. You can't increment it.

But you can set a pointer to the start of the array and then increment
that:

unsigned int *p = &input_0[0]; /* or just p = input */
*p++ = 100;
...

-- Richard

 

[Jack Klein]

Hi,
I have this code,

unsigned int input_0[8];

*input_0++ = 100;
*input_0++ = 200;
*input_0++ = 300;
*input_0++ = 400;
*input_0++ = 1;
*input_0++ = 2;
*input_0++ = 3;
*input_0 = 4;

The compiler is gcc ported to a specific embedded processor. The
compiler complains that I can not use increment operation like that.

Do you know why?

Because input_0 is an array, not a pointer, and you cannot increment
an array.

An array is not a pointer, and a pointer is not an array. Far too
many newbies (and some oldbies) never grasp this.

 

[jaysome]

Hi,
I have this code,

unsigned int input_0[8];

*input_0++ = 100;
*input_0++ = 200;
*input_0++ = 300;
*input_0++ = 400;
*input_0++ = 1;
*input_0++ = 2;
*input_0++ = 3;
*input_0 = 4;

The compiler is gcc ported to a specific embedded processor. The
compiler complains that I can not use increment operation like that.

Do you know why?

Because input_0 is an array, not a pointer, and you cannot increment
an array.

An array is not a pointer, and a pointer is not an array. Far too
many newbies (and some oldbies) never grasp this.

As demonstrated by the following code:

void print_array_size(const int array[])
{
printf("array size = %lu\n", (unsigned long)(sizeof array));
}

int main(void)
{
int array[16] = {0};
print_array_size(array);
return 0;
}

 

[Chris Dollin]

Hi,
I have this code,

unsigned int input_0[8];

*input_0++ = 100;
*input_0++ = 200;
*input_0++ = 300;
*input_0++ = 400;
*input_0++ = 1;
*input_0++ = 2;
*input_0++ = 3;
*input_0 = 4;

The compiler is gcc ported to a specific embedded processor. The
compiler complains that I can not use increment operation like that.

Do you know why?

Yes.

You can't increment an array. Why did you think you could?

 

[Keith Thompson]

[...]

An array is not a pointer, and a pointer is not an array. Far too
many newbies (and some oldbies) never grasp this.

As demonstrated by the following code:

void print_array_size(const int array[])
{
printf("array size = %lu\n", (unsigned long)(sizeof array));
}

int main(void)
{
int array[16] = {0};
print_array_size(array);
return 0;
}

The object "array" in main() is an array.

The object (parameter) "array" in print_array_size() is a pointer, not
an array. The rule that a parameter declaration such as "int foo[]"
is quietly transformed to the equivalent of "int *foo*" is, in my
humble opinion, unfortunate.

 

[jaysome]

[...]

An array is not a pointer, and a pointer is not an array. Far too
many newbies (and some oldbies) never grasp this.

As demonstrated by the following code:

void print_array_size(const int array[])
{
printf("array size = %lu\n", (unsigned long)(sizeof array));
}

int main(void)
{
int array[16] = {0};
print_array_size(array);
return 0;
}

The object "array" in main() is an array.

The object (parameter) "array" in print_array_size() is a pointer, not
an array. The rule that a parameter declaration such as "int foo[]"
is quietly transformed to the equivalent of "int *foo*" is, in my
humble opinion, unfortunate.

Yes. Languages such as C and C++ suffer from this, while languages
such as Ada and C# are more fortunate.

 

[rain.man]

unsigned int input_0[8];
unsigned int *p_inp = input_0;

*p_inp_0++ = 100;

..............................

 

[Charles Richmond]

Hi,
I have this code,

unsigned int input_0[8];

*input_0++ = 100;
*input_0++ = 200;
*input_0++ = 300;
*input_0++ = 400;
*input_0++ = 1;
*input_0++ = 2;
*input_0++ = 3;
*input_0 = 4;

The compiler is gcc ported to a specific embedded processor. The
compiler complains that I can not use increment operation like that.

Do you know why?

Your variable "input_0" is an array name, and thus a constant.
It can *not* be incremented. Try this:

unsigned int *uiray[8];
unsigned int *input_0 = uiray;

*input_0++ = 100;
*input_0++ = 200;
*input_0++ = 300;
*input_0++ = 400;
*input_0++ = 1;
*input_0++ = 2;
*input_0++ = 3;
*input_0 = 4;


In this code, "input_0" is a pointer, *not* an array name.

 

[Scott W]

Hi,
I have this code,

unsigned int input_0[8];

*input_0++ = 100;
*input_0++ = 200;
*input_0++ = 300;
*input_0++ = 400;
*input_0++ = 1;
*input_0++ = 2;
*input_0++ = 3;
*input_0 = 4;

The compiler is gcc ported to a specific embedded processor. The
compiler complains that I can not use increment operation like that.

Do you know why?

Your variable "input_0" is an array name, and thus a constant.
It can *not* be incremented. Try this:

unsigned int *uiray[8];
unsigned int *input_0 = uiray;

*input_0++ = 100;
*input_0++ = 200;
*input_0++ = 300;
*input_0++ = 400;
*input_0++ = 1;
*input_0++ = 2;
*input_0++ = 3;
*input_0 = 4;


In this code, "input_0" is a pointer, *not* an array name.

why not just do:

int a = 0;
unsigned int input_0[8];

input_0[a++] = 100;
input_0[a++] = 200;
input_0[a++] = 300;
input_0[a++] = 400;
input_0[a++] = 1;
input_0[a++] = 2;
input_0[a++] = 3;
input_0[a] = 4;

or hell, just:

unsigned int input_0[8];

input_0[0] = 100;
etc.

 

[Lew Pitcher]

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Hash: SHA1

Hi,
I have this code,

unsigned int input_0[8];

*input_0++ = 100;
*input_0++ = 200;
*input_0++ = 300;
*input_0++ = 400;
*input_0++ = 1;
*input_0++ = 2;
*input_0++ = 3;
*input_0 = 4;

The compiler is gcc ported to a specific embedded processor. The
compiler complains that I can not use increment operation like that.

Do you know why?

Your variable "input_0" is an array name, and thus a constant.
It can *not* be incremented. Try this:

unsigned int *uiray[8];
unsigned int *input_0 = uiray;

*input_0++ = 100;
*input_0++ = 200;
*input_0++ = 300;
*input_0++ = 400;
*input_0++ = 1;
*input_0++ = 2;
*input_0++ = 3;
*input_0 = 4;


In this code, "input_0" is a pointer, *not* an array name.

why not just do:

int a = 0;
unsigned int input_0[8];

input_0[a++] = 100;
input_0[a++] = 200;
input_0[a++] = 300;
input_0[a++] = 400;
input_0[a++] = 1;
input_0[a++] = 2;
input_0[a++] = 3;
input_0[a] = 4;

or hell, just:

unsigned int input_0[8];

input_0[0] = 100;
etc.

or even

int a = 7;
unsigned int input_0[8] = { 100, 200, 300, 400, 1, 2, 3, 4};

which would produce the same end state

- --

Lew Pitcher, IT Specialist, Corporate Technology Solutions,
Enterprise Technology Solutions, TD Bank Financial Group

(Opinions expressed here are my own, not my employer's)
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