[ad_1] setDateTime() accepts ints as parameters, yet you give it char array. The call should be: clock.setDataTime(2017, 3, 22, 16, 20, 4); 0 [ad_2] solved error: invalid conversion from ‘char*’ to ‘uint32_t
[ad_1] I have modified your code. Please see the snipet code below #include <stdio.h> #include <conio.h> #define MAX 25 char welcomeMsg[]= “Please enter your name without ‘*’ or # ” ; char errorMsg[]= “\nError, please re-type your name. “; void main(void) { int j; char input; char name[MAX]; j=0; puts(welcomeMsg); do { input = getche(); … Read more
[ad_1] Use ‘a’ and not “a”, the difference between char and string is subtle here but substantial. [ad_2] solved Error Operator ‘==’ cannot be applied to operands of type ‘char’ and ‘string’ [closed]
[ad_1] EDIT Now, I’ve figured out the correct, working answer based on the article I found. Here is it: function unicode2utf8($str) { $a=json_decode(str_replace(“\n”,”<br>”,'[“‘.$str.'”]’)); return str_replace(“<br>”,”\n”,$a[0]); } PHP: echo unicode2utf8(“Tom\u00e1nsson\n Eriksen\n Gilverto\n”); Output: Tománsson Eriksen Gilverto ORIGINAL I’ve found an article about the same problem here ( http://www.welefen.com/php-unicode-to-utf8.html ) The solution is the following function function … Read more
[ad_1] You can use functions std::stringstream from sstream or strtod from cstdlib or stod (If you are using C++11) as per your need. Quoting example from cplusplus.com // stod example #include <iostream> // std::cout #include <string> // std::string, std::stod int main () { std::string orbits (“365.24 29.53”); std::string::size_type sz; // alias of size_t double earth … Read more
[ad_1] What? A character is typically just that, one character. Not seven. You can do a wide character: char what=”12-4-30″; but I think that’s more bits than most compilers will let you have in a char, so it’s not going to work. If you meant a string of character, you should use snprintf(): char what[32]; … Read more
[ad_1] Here s will be “3xzy1” as s[2]=’x’+2; makes s[2] equal to ‘z’, where ‘z’ is a character, not a string. ‘x’-1==’w’; ‘x’+1==’y’; ‘x’+2==’z’ [ad_2] solved C++ sum char and integer
[ad_1] That looks like a fragment of a JSON document with Cyrillic text. [ad_2] solved How to identify Encoding [closed]
[ad_1] You want a variable lenght array, you can do it like this: int digit; scanf(“%d”, &digit); char test [digit+1]; This will work for your purposes. Usual restrictions of course apply. Keep in mind this functionality was only added in the C99 Standard. So if your compiler supports only previous standards, this will not work. … Read more
[ad_1] First off, you were using the strcpy_s function wrong, you didn’t give the length of the string. So I changed it like; strcpy_s(mask, m, word[x]); And also, the format specifier we’re going to use while getting a string from the user will be %s like; scanf_s(“%s”, &answer); Finally, I prefer using strcmp function to … Read more
[ad_1] Here is a very compact implementation: let to_str_list = List.map (fun cs -> String.concat “” (List.map (String.make 1) cs)) It looks like this when your call it: # to_str_list [[‘h’;’e’;’l’;’l’;’o’];[‘w’;’o’;’r’;’l’;’d’]];; – : string list = [“hello”; “world”] Update If you want to suppress empty strings at the outer level you can do something like … Read more
[ad_1] You are passing a pointer to a local variable. Once your function ends, this variable is gone. If this is C++ you should use the string class. If for whatever reason you don’t, at least be const correct: const char* Worker::getName() const { return name; } 1 [ad_2] solved cout corrupt char* [closed]
[ad_1] I think that you needed string, not 2D-Array. You can be written out to a string like write to the standard output by implementing a string that a simple extension. #include <stdio.h> #include <stdlib.h> #include <string.h> typedef struct char_vec { char *v; size_t capacity; size_t used; } sstream; sstream *ss_new(void); void ss_free(sstream *ss); void … Read more
[ad_1] Try following code: #include<iostream> #include<cstring> using namespace std; #define CHAR_ARRAY_SIZE 8 int main() { char charArray[CHAR_ARRAY_SIZE] = {‘1’, ‘2’, ‘3’, ‘4’, ‘5’, ‘6’, ‘7’, ‘8’}; int intArray[2]; for(int i = 0; i < CHAR_ARRAY_SIZE/4; i++) { char ch1 = (charArray[4*i+0] – ‘0’); char ch2 = (charArray[4*i+1] – ‘0’); char ch3 = (charArray[4*i+2] – ‘0’); … Read more
[ad_1] ‘\060’ is a single character, in which 060 are octal digits, whose value is indeed 48 in decimal. C11 §6.4.4.4 Character constants octal-escape-sequence: \ octal-digit \ octal-digit octal-digit \ octal-digit octal-digit octal-digit [ad_2] solved Why is it possible to store more than one characters in a “char” variable?