You should read the compiler error messages. The second and third arguments to MessageBox have to have the same type. Either you call MessageBoxA with two char *, or you call MessageBoxW with two wchar_t *. One fix for your code would be to do MessageBoxA(NULL, buff, “User-id”, MB_OK). You are using sprintf_s incorrectly too, … Read more
A minimal hello world program in C looks like this : #include <stdio.h> int main() { printf(“Hello World\n”); return 0; } You first have to include stdio.h which gives you access to the printf function. Then you have to define a main function which is the entry point of the program. The code inside the … Read more
printf without arguments is the function pointer, worth a non-zero value (it’s built-in so the pointer cannot be zero) Now you apply logical negation (!) on this non-zero value: you get zero. Now negate this zero bit-wise (using bit to bit negation ~), you get all 1s in the int bit range (which can vary … Read more
Your message functions lack a return type. C deduces it to be int. However, not returning a value from a non-void function is Undefined Behaviour. In your case, the return value from printf has probably been stored in a register, which was not overwritten by message before it itself returned. Thus, the return value propagated. … Read more
As you have not posted code, I will also just verbally describe a possible algorithm. You’ll need two embedded for loops. Outside loop for lines (y), inside loop for characters within line (x). Both shall run from zero to xmax and ymax, which are the numbers specified on the command line minus one. Inside the … Read more
In your: if ( buff == ‘\n’ && fe && fn ) //<– I assume you want to check if an empty line was read by checking it against the newline character. However, buff is not the character; it is a pointer to were all the characters read are stored. To check if the first … Read more
.7 is the precision and le means normal form. double x = 0.012345678910; printf(“%.7lf\n”, x); // 0.0123457 printf(“%.7le\n”, x); // 1.2345679e-02 printf(“%.9le\n”, x); // 1.234567891e-02 Edit See Eric Postpischil’s unswer below, it’s way more comprehensive than this one. 2 solved What does “%.7le ” in printf means?
For size_t, assuming you have a sufficiently modern C library, use %zu. If you can’t use the z modifier (some older libraries unfortunately don’t support it), cast to a wide-enough known type when printing, and then use a width specifier appropriate to that type: size_t sz = sizeof(whatever); … printf(“%lu”, (unsigned long)sz); This works as … Read more
#include <stdio.h> #include <stdlib.h> int parabola1(int); int *calc(int low, int high, int (*f)(int), int *size, int *min, int *max){ /* #input low, high : range {x| low <= x <= high} f : function #output *size : Size of array *min : Minimum value of f(x) *max : Maximum value return : pointer to first … Read more
Why here the output is only “hie” and “hola”? Order of precedence of Logical AND (&&) is greater than Logical OR (||). Agreed. But, it doesn’t mean that a program has to evaluate in that order. It just says to group together the expressions. Hence, if(printf(“hie”)|| printf(“hello”)&& printf(“nice to see you”)) is equivalent to, if(printf(“hie”) … Read more
Characters are encoded as integers, usually 8-bit by default in C. Some are defined by ASCII and others depend on your OS, screen font, etc. The * character has code 42. If you enter 55, your code computes 42*55=2310 and uses the low 8 bits of this value, which is 6, as the character to … Read more
example by use strtok, strchr, sprintf #include <stdio.h> #include <stdlib.h> #include <string.h> int main(){ const char *data = “date=2013-12-09 time=07:31:10 d_id=device1 logid=01 user=user1 lip=1.1.1.1 mac=00:11:22:33:44:55 cip=2.2.2.2 dip=3.3.3.3 proto=AA sport=22 dport=11 in_1=eth1 out_1=”; char *work = strdup(data);//make copy for work char *output = strdup(data);//allocate for output char *assignment; //tokenize to aaa=vvv size_t o_count = 0;//output number … Read more
I think what you are missing is, your one line with nested function calls is basically same as this code: int scanf_result = scanf(“%d%d,”,&i,&a); int printf_result = printf(“PRINT %d\t”, scanf_result)); printf(“%d”, printf_result); scanf call should return 2 if you entered valid input, but can also return -1 for actual error, or 0 or 1 if … Read more
The program most likely gets stuck on the system call. You are printing out text without the trailing newline, so it will be kept in the buffer, until a newline is printed, which you do after the system call. 5 solved While true not working [duplicate]
all you have to do is correct the %l(there is not such a specifier), you probably should have used %ld. char *buildStatus() { struct rusage *usage = malloc(sizeof(struct rusage)); int usageRet = getrusage(RUSAGE_SELF, usage); if (usageRet == -1) { perror(“RUSAGE fail”); exit(EXIT_FAILURE); } long unsigned cpuTime = (usage->ru_utime).tv_sec + (usage->ru_stime).tv_sec; long memUsed = get_memory_usage_linux(); unsigned … Read more