C Print Full Data Of Char Pointer Stack Overflow

C Print Full Data Of Char Pointer Stack Overflow I'm new in c and i want to show data of char* pointer before writing it to file but the output only 5 first bytes. i don't know what is wrong with it. i'm using android log print to show log. t. I'm currently studying c and i'm trying to just print the contents of a string array. i'm using pnames to point to the first char pointer and iterating from there. a more proper approach would use this pointer, get a char* each time and use printf("%s", pnames[i]) to print a whole string.

C Print Full Data Of Char Pointer Stack Overflow The %c format specifier expects a char type, and will output a single char value. the first parameter to printf must be a const char* (a char* can convert implicitly to a const char*) and points to the start of a string of characters. 1) char *p1; declares a variable (p1) that is a pointer. the pointer, p1 can be changed. 2) char *p1 = "abcd"; declares p1 (same as "1") and also initializes the data it points to (initializes it to the read only, 5 lelement character array "abcd\0"). So a possible solution is this code, using an additional variable: char *arr = "this is to test"; char c = *( arr); printf("\n%c %c ", c, c); return 0; in this code, arr will only be incremented once and the character fetched at this position can be used later. arr is equivalent to arr = 1. you are actually incrementing arr. It's the %s specifier for a string of characters, which stops printing when it reaches a null character (\0). %u on the other hand, is the specifier for an unsigned decimal integer.

Print Char Pointer Value In C Stack Overflow So a possible solution is this code, using an additional variable: char *arr = "this is to test"; char c = *( arr); printf("\n%c %c ", c, c); return 0; in this code, arr will only be incremented once and the character fetched at this position can be used later. arr is equivalent to arr = 1. you are actually incrementing arr. It's the %s specifier for a string of characters, which stops printing when it reaches a null character (\0). %u on the other hand, is the specifier for an unsigned decimal integer. Char* is just a pointer that points to the beginning of the string. many c functions (printf, strcpy, strlen, ) depend on the terminating '\0' at the end of the string into which a passed to them pointer happens to point to. Scanf( "%s", str ); printf( "%s", str ); you don't need the & for either call. the %s conversion specifier expects the corresponding argument to have type char *. under most circumstances 1, an expression of array type is converted to an expression of pointer type, and the value of the expression is the address of the first element in the array. Here, char* s looks like it contains letters, not an address. as explained above, it doesn't actually. thanks to the syntactic sugar, some actual work is done for you underneath the hood that ends up storing the string somewhere in memory and storing the address of that memory block into s. Pointed to by the pointer and this is the c character. when you do cout << pname you are passing the pointer itself. and cin will print out the whole string. in other words it would do cout << *pname, then cout << * (pname 1), until it has stepped through the whole string. thank for the answer guestgulkan i know that *pname will print c.