我收到这组gcc错误,我似乎无法从谷歌或手册页得到答案.任何洞察这些意味着什么或从哪里开始寻找?
这是makefile中的行:
#After building several otehr bits of code into static libraries
...
# Grand finally link all the object files into one
gcc --shared \
-m64 \
-Wl,--whole-archive ./release64/*.a \
-o ./release64/libMYLIB.so.1.0
ln -sf libArcGIS.so.1.0 ./release64/libMYLIB.so
ln -sf libArcGIS.so.1.0 ./release64/libMYLIB.so.1
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我得到了以下错误(还有更多我采取了前n只是为了给出一个想法:
/usr/lib/gcc/x86_64-redhat-linux/4.4.5/libgcc.a(_muldi3.o): In function `__multi3':
(.text+0x0): multiple definition of `__multi3'
/usr/lib/gcc/x86_64-redhat-linux/4.4.5/libgcc.a(_muldi3.o):(.text+0x0): first defined here
/usr/lib/gcc/x86_64-redhat-linux/4.4.5/libgcc.a(_negdi2.o): In function `__negti2':
(.text+0x0): multiple definition of `__negti2'
/usr/lib/gcc/x86_64-redhat-linux/4.4.5/libgcc.a(_negdi2.o):(.text+0x0): first defined here
/usr/lib/gcc/x86_64-redhat-linux/4.4.5/libgcc.a(_lshrdi3.o): In function `__lshrti3':
(.text+0x0): multiple definition of `__lshrti3'
/usr/lib/gcc/x86_64-redhat-linux/4.4.5/libgcc.a(_lshrdi3.o):(.text+0x0): first defined …Run Code Online (Sandbox Code Playgroud) 我希望通过这个小脚本总结出我想要实现的目标.
DIRS = dir1 dir2 dir3 dir4 ...
OBJS =
all: GENERATE_OBJECT_FILES
GENERATE_OBJECT_FILES:
for curr_dir in $(DIRS); \
do \
$(join $(OBJS), `ls $${curr_dir}/*.o`); \
done
echo $(OBJS);
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如何使用Makefile中的脚本完成此操作?
我在用C编写的程序遇到了一些问题,而且我已经超越了我的知识.总之,我需要将链接列表从一个列表深层复制到另一个列表.这些列表中包含malloc数据,我需要保留所有数据,而不需要指针指向相同的信息.
我只发布了我认为相关的代码.如果我遗漏了任何重要的背景信息,请告诉我.
这是代码:
typedef struct matrix {
char *name;
int R;
int C;
int dim;
void (*concat_matrices)( struct matrix *A, struct matrix *B, struct matrix *ret );
} Matrix;
void concat_matrices( Matrix *A, Matrix *B, Matrix *ret ) {
int L1 = strlen( A->name );
int L2 = strlen( B->name );
int len = L1 + L2;
char *Ap = (char*)malloc(L1*sizeof(char)); strcpy(Ap,A->name);
char *Bp = (char*)malloc(L2*sizeof(char )); strcpy(Bp,B->name);
char *c = (char*)malloc(sizeof(char)*(len + 2));
c[0] = '('; strcat(c, Ap); …Run Code Online (Sandbox Code Playgroud) 所以我有一个简单的C程序循环传递给main的args然后返回:
#include <stdio.h>
int main(int argc, char *argv[])
{
int i;
for(i = 0; i < argc; ++i) {
fprintf(stdout, "%s\n", argv[i]);
}
return 0;
}
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我想看看gcc如何用NASM格式写出程序集.我正在查看.asm文件中的输出,并注意到语法是TASM.下面是make文件和gcc的输出.我做错了还是gcc没有输出真正的NASM语法?
all: main
main: main.o
ld -o main main.o
main.o : main.c
gcc -S -masm=intel -o main.asm main.c
nasm -f elf -g -F stabs main.asm -l main.lst
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和
.file "main.c"
.intel_syntax noprefix
.section .rodata
.LC0:
.string "%s\n"
.text
.globl main
.type main, @function
main:
push ebp
mov ebp, esp
and esp, -16
sub esp, …Run Code Online (Sandbox Code Playgroud) 在我努力了解如何使用 GNU binutils 使用 Gas构建简单的引导加载程序时,我遇到了一个问题,如何告诉链接器将数据放在使用 .org 推进位置计数器的文件中的位置同时将文件大小保持在 512 字节。我似乎找不到办法做到这一点。
尝试执行此操作的汇编程序是:
# Author: Matthew Hoggan
# Date Created: Tuesday, Mar 6, 2012
.code16 # Tell assembler to work in 16 bit mode
.section .data
msg: # Message to be printed to the screen
.asciz "hello, world"
.section .text
.globl _start # Help linker find start of program
_start:
xor %ax, %ax # Zero out ax register (ah used to specify bios function to Video Services)
movw %ax, %ds # Since …Run Code Online (Sandbox Code Playgroud) 我在接受采访时被问到这个问题,但仍无法弄清楚它的作用.有人可以解释它的作用以及它是如何做到的吗?
v = v - ((v >> 1) & (T)~(T)0/3); // temp
v = (v & (T)~(T)0/15*3) + ((v >> 2) & (T)~(T)0/15*3); // temp
v = (v + (v >> 4)) & (T)~(T)0/255*15; // temp
c = (T)(v * ((T)~(T)0/255)) >> (sizeof(T) - 1) * CHAR_BIT; // count
Run Code Online (Sandbox Code Playgroud) 我的嵌套意图服务定义如下:
package com.my.package;
... // Bunch of imports
public class MyNotifier
... // Bunch of variables
public class MissedCallIntentService extends IntentService {
private static final String TAG = "MissedCallIntentService";
public MissedCallIntentService() {
super("MissedCallIntentService");
Log.i(TAG, "Creating intent service.");
}
@Override
public void onHandleIntent(Intent intent) {
Log.i(TAG, "Handling intent service.");
}
}
// Test my nested intent filter
public MyNotifier(Context app) {
mApp = app;
Log.i(LOG_TAG, "Going to start intent service.");
Intent intent = new Intent(mApp, MissedCallIntentService.class);
mApp.startService(intent);
}
... // Bunch …Run Code Online (Sandbox Code Playgroud) 我有这两个文件 myboot.asm 和 Theirboot.asm (分别列出):
;----------------------------------------------------------------------
; A Simple boot program that prints the string 'Hello World'
; Author: Matthew Hoggan 2012
;----------------------------------------------------------------------
Output db 'Hello',0x00 ; Output string for bios
org 0x7c00 ; This is where BIOS loads the bootloader
entry: ; Label to Signify entry to program
jmp short begin ; Jump over the DOS boot record data
; --------------------------------------------
; Boot program code begins here
; --------------------------------------------
begin: ; Label to Signify entry to program
mov si, …Run Code Online (Sandbox Code Playgroud) 要查看我要执行的操作,请参见以下内容:
我的问题是如何在我的Makefile.am中有条件地设置AM_CPPFLAGS或my_lib_la_CPPFLAGS。这样在运行configure时是否设置了正确的CPPFLAGS?
目前,我正在做一些影响以下方面的事情:
lib_xml_wrapper_la_CPPFLAGS = -I../../
UNAME_S = $(shell uname -s)
UNAME_P = $(shell uname -p)
ifeq ($(UNAME_S),Linux)
lib_xml_wrapper_la_CPPFLAGS += -DLINUX
ifeq ($(UNAME_P),x86_64)
lib_xml_wrapper_la_CPPFLAGS += -D AMD64
endif
ifeq ($(UNAME_P),x86_64)
lib_xml_wrapper_la_CPPFLAGS += -I../../../external/xerces-c-3.1.1-x86_64-linux-gcc-3.4/include/
endif
ifneq ($(filter %86,$(UNAME_P)),)
lib_xml_wrapper_la_CPPFLAGS += -I../../../external/xerces-c-3.1.1-x86-linux-gcc-3.4/include/
endif
ifneq ($(filter arm%,$(UNAME_P)),)
lib_xml_wrapper_la_CPPFLAGS +=
endif
endif
ifeq ($(UNAME_S),Darwin)
lib_xml_wrapper_la_CPPFLAGS += -DOSX
ifneq ($(filter %86,$(UNAME_P)),)
lib_xml_wrapper_la_CPPFLAGS += -I../../../external/xerces-c-3.1.1-x86-macosx-gcc-3.4/include/
endif
ifneq ($(filter arm%,$(UNAME_P)),)
lib_xml_wrapper_la_CPPFLAGS +=
endif
endif
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这似乎在Makefile.am中不起作用。我收到以下错误:
xml_wrapper/Makefile.am:26: error: endif without if
xml_wrapper/Makefile.am:35: error: endif without if
automake: warnings …Run Code Online (Sandbox Code Playgroud) 给定此顶点着色器:
attribute vec3 vertex;
uniform mat4 mvp;
void main() {
gl_Position = mvp * vec4(vertex, 1.0);
}
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这个片段着色器:
void main() {
gl_FragColor = vec4(1.0, 0.0, 0.0, 1.0);
}
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当mvp矩阵是标识或模型矩阵是缩放,旋转或转换变换时,哪个能够渲染下面的数据:
GLfloat values[] = {
-1.0, -1.0, +0.0,
+1.0, -1.0, +0.0,
+0.0, +1.0, +0.0,
};
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为什么以下使用Qt QMatrix4x4::lookAt并QMatrix4x4::perspective导致场景被渲染,好像没有对象?
QMatrix4x4 model;
QMatrix4x4 view;
view.lookAt(
QVector3D(0.0, 0.0, 10.0), // Eye
QVector3D(0.0, 0.0, 0.0), // Focal Point
QVector3D(0.0, 1.0, 0.0)); // Up vector
QMatrix4x4 proj;
// Window size is fixed at …Run Code Online (Sandbox Code Playgroud)