Use a conditional breakpoint that checks the first parameter. On 64-bit x86 systems the condition would be: (gdb) b write if 1==$rdi On 32-bit systems, it is more complex because the parameter is on the stack, meaning that you need to cast $esp to an int * and index the fd parameter. The stack at … Read more
The PRIuPTR macro (from <inttypes.h>) defines a decimal format for uintptr_t, which should always be large enough that you can cast a size_t to it without truncating, e.g. fprintf(stream, “Your size_t var has value %” PRIuPTR “.”, (uintptr_t) your_var);
You can find a list of formatting escapes on this page. %d is a signed integer, while %u is an unsigned integer. Pointers (when treated as numbers) are usually non-negative. If you actually want to display a pointer, use the %p format specifier.
You can do it with 1 printf: System.out.printf(“First Name: %s\nLast Name: %s”,firstname, lastname);
This prints out all ASCII values: int main() { int i; i=0; do { printf(“%d %c \n”,i,i); i++; } while(i<=255); return 0; } and this prints out the ASCII value for a given character: int main() { int e; char ch; clrscr(); printf(“\n Enter a character : “); scanf(“%c”,&ch); e=ch; printf(“\n The ASCII value of … Read more
Usually you can use a cast to convert the operand to some type for which you know the right format. Your proposed solution: time_t t = time(NULL); printf(“%s”, t); clearly will not work, since time_t is a numeric type, not char*. We know, in general, that time_t is an arithmetic type. Something like this: printf(“%ld\n”, … Read more
They are completely equivalent when used with printf(). Personally, I prefer %d. It’s used more often (should I say “it’s the idiomatic conversion specifier for int“?). (One difference between %i and %d is that when used with scanf(), then %d always expects a decimal integer, whereas %i recognizes the 0 and 0x prefixes as octal … Read more
If you try this: #include<stdio.h> void main() { char name[]=”siva”; printf(“name = %p\n”, name); printf(“&name[0] = %p\n”, &name[0]); printf(“name printed as %%s is %s\n”,name); printf(“*name = %c\n”,*name); printf(“name[0] = %c\n”, name[0]); } Output is: name = 0xbff5391b &name[0] = 0xbff5391b name printed as %s is siva *name = s name[0] = s So ‘name’ is … Read more
You had originally tagged this C++, but it would appear to be a C program. C will automatically provide an implicit declaration for a function if there is no prototype in scope (such as due to the omission of #include <stdio.h>). The implicit declaration would be: int printf(); Meaning that printf is a function that … Read more