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No, array-type is implicitly converted into pointer type when you pass it in to a function.

So,

void PrintSize(int p_someArray[10]) { printf("%zu\n", sizeof(p_someArray)); } 

and

void PrintSize(int *p_someArray) { printf("%zu\n", sizeof(p_someArray)); } 

are equivalent. So what you get is the value of sizeof(int*)

An array-type is implicitly converted into pointer type when you pass it in to a function.

So,

void PrintSize(int p_someArray[10]) { printf("%zu\n", sizeof(p_someArray)); } 

and

void PrintSize(int *p_someArray) { printf("%zu\n", sizeof(p_someArray)); } 

are equivalent. So what you get is the value of sizeof(int*)

No, array-type is implicitly converted into pointer type when you pass it in to a function.

So,

void PrintSize(int p_someArray[10]){ printf("%zu\n", sizeof(p_someArray)); } 

and

void PrintSize(int *p_someArray){ printf("%zu\n", sizeof(p_someArray)); } 

are equivalent. So what you get is the value of sizeof(int*)

No, array-type is implicitly converted into pointer type when you pass it in to a function.

So,

void PrintSize(int p_someArray[10]) {  printf("%zu\n", sizeof(p_someArray)); } 

and

void PrintSize(int *p_someArray) {  printf("%zu\n", sizeof(p_someArray)); } 

are equivalent. So what you get is the value of sizeof(int*)

No, an array-type is implicitly converted into pointer type when you pass it in to a function.

So,

void PrintSize(int p_someArray[10]){ printf("%zu\n", sizeof(p_someArray)); } 

and

void PrintSize(int *p_someArray){ printf("%zu\n", sizeof(p_someArray)); } 

are equivalent. So what you get is the value of sizeof(int*)

No, array-type is implicitly converted into pointer type when you pass it in to a function.

So,

void PrintSize(int p_someArray[10]){ printf("%zu\n", sizeof(p_someArray)); } 

and

void PrintSize(int *p_someArray){ printf("%zu\n", sizeof(p_someArray)); } 

are equivalent. So what you get is the value of sizeof(int*)