unlike suggested in the comments to the question, sint32 is not uncommon in safety critical and embedded systems and usually falls back to a typecast for int. (e.g. in some MISRA environments). Hence sint32 a = -1; printf(“%d”, a); should do the trick anyways. tested with gcc v5.2.1 and arm-gcc v5.2.1 (-Wall and no warnings). … Read more
Make sure that Cards::init() is called and returns true before calling Cards::load. The array accessed by Cards::load will consist of three empty strings in case Cards::init() is not called, or if it returns false. 1 solved How to print std::string array with printf in c++? [closed]
The error is that %d in the format specifier of both scanf and printf represents decimal, so it expects an int (in the case of scanf, a pointer to int) Since you are declaring b and c as float, %d in the lines scanf(“%d”,&b); printf(“%d lbs is %d kgs.”,b,c); should be changed to %f respectively. … Read more
The answer is a multi-byte character that is an bug from the runtime. You can work around it by finding chr(16) in your string and replacing it with an empty string (“”). That is if you need it to work with s/v/printf functions. solved PHP printf not printing, simplexml_load_string not parsing [closed]
There is one semicolon after the if statement that is causing the problem. if (i!=0); { //this will always execute } change it to if (i!=0) { //this will execute if i != 0 } The compiler does not warn you because the first statement is syntactically valid. solved C: Weird conditional printf behavior [closed]
You may be able to see what’s happening more clearly if you split the statement into several statements: int temp1 = printf(“Hello world!\n”); int temp2 = printf(“%d”, temp1); printf(“%d”, temp2); The first printf prints Hello world!\n. Since this is 13 characters, it returns 13. The second printf prints 13. Since this is 2 characters, it … Read more
i intentionally left the previous answer because understanding memory allocation is trivial in programming in c specially. and as i see you have a big issue with that. but still you have issue in nearly every thing. in my actual answer, i will try to simplify you how to use strtok, to split string and … Read more
%x format specifier experts the argument to be of type unsigned int. In your case, printf(“%x\n”,(const uint8_t)0x0D); printf(“%x\n”,0x0D); arguments will be promoted (default promotion rule) to match the type, but in case of printf(“%x\n”,(const uint8_t *)0x0D); //supplying a pointer printf(“%x\n”,(uint8_t *)0x0D); //supplying a pointer You’ll invoke undefined behavior, as per C11, chapter §7.21.6.1 […] If … Read more
Variable base is undeclared. Variable idade is undeclared. It should be printf(“%d”,base) if your are using base. #include <stdio.h> int main(){ int age; int aget; int total; printf(“Input your age: “); scanf(“%i”, &age); printf(“Input your age two: “); scanf(“%i”, &aget); total = age + aget; printf(“%d”,total); return 0; } 1 solved How to solve segmentation … Read more
As for me then this return 0*printf(“%d”,a[i]); just a bad programming style. At least it would be better to write instead return ( printf(“%d”,a[i]), 0 ); not saying about printf(“%d”,a[i]); return 0; Maybe this statement is found in a recursive function. As for your question Having trouble understanding output of line of code – then … Read more
You misunderstood how printf works. It does not concatenate all the stuff you throw into it like you’re used to from std::cout << … << …; but instead mostly prints the first argument, the format string. The latter arguments only come into play when the format string requests them, as you do with the %0.4f … Read more
These are the very basics of C Programming, and I strongly advise you to get a decent book – The C Programming Language by Dennis Ritchie would be a good start. There are numerous errors in your code. A char can contain only one character, like ‘A’, or ‘a’ or something like that. When you’re … Read more
Try this: sub(/^0+/, “”, $2)? $ awk ‘BEGIN{b=”000000000000787301″; sub(/^0+/, “”, b); print b;}’ 787301 or typecast it by adding zero: $ awk ‘BEGIN{b=”000000000000787301″; print b+0;}’ 787301 update A working example based on comments: $ echo ‘E2EDP19001 02002000000000000797578’ | awk ‘{$2 = substr($2,6); sub(/^0+/, “”, $2); print $2;}’ 797578 or, preserve $2: $ echo ‘E2EDP19001 02002000000000000797578’ … Read more
It’s the oldest question ever. Why do people find this so fascinating? This is undefined behavior; you’re relying on side-effects without a sequence point between modifications. solved ambigious behaviour of increment operator in printf [duplicate]
I got it. I wrote: echo printf(…); instead of just: printf(…); After string produced by printf() i got that string length, which is value returned by printf (in fact – added number was 9 in case of 123.45 value, in 20.4 case it was number 8). 2 solved PHP printf adds something after formatted string