C - Pointer inside a function pointing to an array displaying invalid character

Question

I have an array product[6] and a pointer (*productPtr) pointing to this array. When I pass the pointer as an argument to a function called productCheck and try to print one of the characters, I get an invalid reading of the characters from the product array. Any help as to why this is happening is greatly appreciated. For example if i = 3, instead of reading 'u', the output is '{'.

int i = 3;    
char product[6] = "xddua";
char * productPtr;
productPtr = product;


bool productCheck(int i, char * productPtr);


productOk = productCheck(i, &product);


bool productCheck(int i, char * productPtr)
{    
    printf("product is %c\n", *productPtr + i);

*productPtr + i-> *(productPtr + i) or arguably more clearly/simply productPtr[i] — kaylum
– kaylum, Commented May 1, 2021 at 8:09
regarding: productOk = productCheck(i, &product); product (the second parameter) is a 'bare name of an array' so will be the address of the first byte of the array, this statement is trying to take the address of the address of the array. Suggest removing the & before produce — user3629249
– user3629249, Commented May 2, 2021 at 4:33
this is nothing but a code fragment. Please post a minimal reproducible example so we can reproduce the problem so we can help you debug it. — user3629249
– user3629249, Commented May 2, 2021 at 4:34
regarding: char * productPtr; productPtr = product; This pair of statements are not used so can be removed — user3629249
– user3629249, Commented May 2, 2021 at 4:37

Vlad from Moscow · Accepted Answer · 2021-05-01 08:32:53Z

2

The function has the second parameter of the type char *.

bool productCheck(int i, char * productPtr);

You are calling the function passing expression the expression &product as the second argument

productOk = productCheck(i, &product);

As the array product is declared like

char product[6] = "xddua";

then the type of the expression ^product is char ( * )[6]. That is the type of the argument is not compatible with the type of the function parameter.

You need to call the function either like

productOk = productCheck(i, product);

or like

productOk = productCheck(i, productPtr);

Within the function dereferencing the passed value of the expression &product in this statement

printf("product is %c\n", *productPtr + i);

And then to the result expression you are adding the value i instead of at first to add the value i to the pointer and then to dereference the pointer expression..

Thus you need to call the function like

productOk = productCheck(i, product);

or like

productOk = productCheck(i, productPtr);

and within the function you need to write either

printf("product is %c\n", *( productPtr + i ) );

or

printf("product is %c\n", productPtr[i] );

edited May 1, 2021 at 8:32

answered May 1, 2021 at 8:22

Vlad from Moscow

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Black Solis Over a year ago

Thanks for all the info! Got it working from this and another answer.

MikeCAT · Accepted Answer · 2021-05-01 08:10:48Z

1

The unary * operator has higher precedence than + operator.

*productPtr + i first reads what is pointed at by productPtr, then adds i to the value read.

To access another element, You should write *(producePtr + i) or productPtr[i].

Also the argument &product is not good. Most arrays in expressions are automatically converted to pointers to the first element (one of the exceptions is the operand of unary &), so it should be product to pass char* value.

answered May 1, 2021 at 8:10

MikeCAT

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1 Comment

Black Solis Over a year ago

Thanks for all the info! Got it working from this and another answer.

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