Remove trailing whitespace

4 files changed, 9 insertions, 9 deletions

diff --git a/common/canvas_utils.c b/common/canvas_utils.c
index 46eb012..a3e205f 100644
--- a/common/canvas_utils.c
+++ b/common/canvas_utils.c
@@ -307,7 +307,7 @@ pixman_image_t *alloc_lz_image_surface(LzDecodeUsrData *canvas_data,
 
     /* pixman requires strides to be 4-byte aligned */
     stride = SPICE_ALIGN(stride, 4);
-    
+
     if (!top_down) {
         stride = -stride;
     }
diff --git a/common/lines.c b/common/lines.c
index 36b40a7..10ca318 100644
--- a/common/lines.c
+++ b/common/lines.c
@@ -927,7 +927,7 @@ end of the line, we will find the largest number of Y steps that
 satisfies the inequality.  In that case, since we are representing
 the Y steps as (dy - N), we will actually want to solve for the
 smallest N in that equation.
-
+
 Case 1:  X major, starting X coordinate moved by M steps
 
                 -2dx <= 2Mdy - 2Ndx - dx - B < 0
@@ -975,7 +975,7 @@ steps, so we want the highest N, so we use the < inequality:
           = floor((2Mdy + dx + B + 2dx - 1) / 2dx) - 1
           = floor((2Mdy + dx + B + 2dx - 1 - 2dx) / 2dx)
           = floor((2Mdy + dx + B - 1) / 2dx)
-
+
 Case 3: Y major, starting X coordinate moved by M steps
 
                 -2dy <= 2Ndx - 2Mdy - dy - B < 0
@@ -1021,7 +1021,7 @@ Same analysis as Case 4, but we want the smallest number of Y steps
 which means the largest N, so we use the <= inequality:
 
         N = floor((2Mdy + dy - B) / 2dx)
-
+
 Now let's try the Y coordinates, we have the same 4 cases.
 
 Case 5: X major, starting Y coordinate moved by N steps
@@ -1066,7 +1066,7 @@ Same derivations as Case 6, but we want the smallest # of X steps
 which means the largest M, so use the <= inequality:
 
         M = floor((2Ndx + dx - B) / 2dy)
-
+
 Case 7: Y major, starting Y coordinate moved by N steps
 
                 -2dy <= 2Ndx - 2Mdy - dy - B < 0
@@ -1111,7 +1111,7 @@ steps which means the largest M, so we use the < inequality:
           = floor((2Ndx + dy + B + 2dy - 1) / 2dy) - 1
           = floor((2Ndx + dy + B + 2dy - 1 - 2dy) / 2dy)
           = floor((2Ndx + dy + B - 1) / 2dy)
-
+
 So, our equations are:
 
         1:  X major move x1 to x1+M     floor((2Mdy + dx - B) / 2dx)
diff --git a/common/log.c b/common/log.c
index fc5c129..df394d2 100644
--- a/common/log.c
+++ b/common/log.c
@@ -52,7 +52,7 @@ static const char * spice_log_level_to_string(SpiceLogLevel level)
     };
 #endif
     const char *str = NULL;
- 
+
     if (level < SPICE_N_ELEMENTS(to_string)) {
         str = to_string[level];
     }
@@ -76,7 +76,7 @@ void spice_logv(const char *log_domain,
                 va_list args)
 {
     const char *level = spice_log_level_to_string(log_level);
-    
+
     if (debug_level == -1) {
         debug_level = getenv("SPICE_DEBUG_LEVEL") ? atoi(getenv("SPICE_DEBUG_LEVEL")) : SPICE_LOG_LEVEL_WARNING;
     }
diff --git a/common/sw_canvas.c b/common/sw_canvas.c
index 7d67ca5..a43cee4 100644
--- a/common/sw_canvas.c
+++ b/common/sw_canvas.c
@@ -95,7 +95,7 @@ static pixman_image_t *get_image(SpiceCanvas *canvas, int force_opaque)
         uint32_t *data;
         int stride;
         int width, height;
-        
+
         /* Remove alpha bits from format */
         format = (pixman_format_code_t)(((uint32_t)format) & ~(0xf << 12));
         data = pixman_image_get_data (sw_canvas->image);