printf

问题 here is a demo.cu aiming to printf from the GPU device: #include "cuda_runtime.h" #include "device_launch_parameters.h" #include <stdio.h> __global__ void hello_cuda() { printf("hello from GPU\n"); } int main() { printf("hello from CPU\n"); hello_cuda <<<1, 1>>> (); cudaDeviceSynchronize(); cudaDeviceReset(); printf("bye bye from CPU\n"); return 0; } it compiles and runs: $ nvcc demo.cu $ ./a.out that's the output that I get: hello from CPU bye bye from CPU Q: why there is no printing result

问题 I'm programming with gcc in CentOS 5.5 and the most of time I use printf() and fprintf() to print on terminal, but in some websites I've seen that some people use write() . I want to know if there's other ways to print on terminal. Thanks. 回答1: There are some major differences between these functions. The standard library provides some functions to output to stdout : printf , puts , putchar etc. And some functions to output to a stream, you can specify the stream to stdout : fprintf , fputs ,

问题 I'm programming with gcc in CentOS 5.5 and the most of time I use printf() and fprintf() to print on terminal, but in some websites I've seen that some people use write() . I want to know if there's other ways to print on terminal. Thanks. 回答1: There are some major differences between these functions. The standard library provides some functions to output to stdout : printf , puts , putchar etc. And some functions to output to a stream, you can specify the stream to stdout : fprintf , fputs ,

问题 This question already has answers here : Closed 8 years ago . Possible Duplicate: mixing cout and printf for faster output I'm using Microsoft Visual Studio 6.0. The following program, #include "stdafx.h" #include "iostream.h" int main(int argc, char* argv[]) { printf("a"); printf("b"); printf("c"); return 0; } produces "abc". While the following program, #include "stdafx.h" #include "iostream.h" int main(int argc, char* argv[]) { printf("a"); cout<<"b"; printf("c"); return 0; } produces "acb

问题 Assuming the following: sizeof(char) = 1 sizeof(short) = 2 sizeof(int) = 4 sizeof(long) = 8 The printf format for a 2 byte signed number is %hd , for a 4 byte signed number is %d , for an 8 byte signed number is %ld , but what is the correct format for a 1 byte signed number? 回答1: what is the correct format for a 1 byte signed number? %hh and the integer conversion specifier of your choice (for example, %02hhX . See the C11 standard, §7.21.6.1p5: hh Specifies that a following d , i , o , u ,

问题 Assuming the following: sizeof(char) = 1 sizeof(short) = 2 sizeof(int) = 4 sizeof(long) = 8 The printf format for a 2 byte signed number is %hd , for a 4 byte signed number is %d , for an 8 byte signed number is %ld , but what is the correct format for a 1 byte signed number? 回答1: what is the correct format for a 1 byte signed number? %hh and the integer conversion specifier of your choice (for example, %02hhX . See the C11 standard, §7.21.6.1p5: hh Specifies that a following d , i , o , u ,

问题 It seems that I can't call sprintf() correctly in assembly. When I try to dprintf() my buffer that should now be formatted, all I get is: (null) and a segmentation fault. When running lldb with my program, strlen() is the reason of the fail as it can't find a \0 in my buffer. Here's my code: mov rdi, buff mov rsi, 0 mov rdx, 17 call memset lea rsi, [rel n_head] mov rdx, rax call sprintf mov rdx, rdi lea rsi, [rel fmt] mov rdi, 1 call dprintf ... section .data n_head: db "Low battery: %d%%", 0

问题 I am rewriting this post because I managed to figure out the problem. The problem with my extremely broken output was due to an improper dynamic memory allocation. Basically I needed to allocate memory for an array of pointers that pointed to structs, but the array itself was nested inside of another struct and the nesting confused me slightly and I ended up over complicating it. So I had a struct named Catalog, that my array was in and that array pointed to another struct named Books. When I

问题 Here's what libc has to say about variadic functions: Since the prototype doesn’t specify types for optional arguments, in a call to a variadic function the default argument promotions are performed on the optional argument values. This means the objects of type char or short int (whether signed or not) are promoted to either int or unsigned int, as appropriate; and that objects of type float are promoted to type double. So, if the caller passes a char as an optional argument, it is promoted

问题 This question already has answers here : How do you allow spaces to be entered using scanf? (11 answers) Closed last year . #include <stdio.h> #define MAX_STRING_LENGTH 1024 int main(){ char input_name_string[MAX_STRING_LENGTH+1],motive_string[MAX_STRING_LENGTH+1]; printf("What is your name?\n"); scanf("%1024s",input_name_string); printf("your name is %s \n", input_name_string); printf("What is your motive?\n"); scanf(" %1024s",motive_string); printf("your motive is %s \n", motive_string);